# Minimum number of jumps to reach end

• Difficulty Level : Medium
• Last Updated : 28 Sep, 2022

Given an array arr[] where each element represents the max number of steps that can be made forward from that index. The task is to find the minimum number of jumps to reach the end of the array starting from index 0. If the end isn’t reachable, return -1.

Examples:

Input: arr[] = {1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9}
Output: 3 (1-> 3 -> 9 -> 9)
Explanation: Jump from 1st element to 2nd element as there is only 1 step.
Now there are three options 5, 8 or 9. I
f 8 or 9 is chosen then the end node 9 can be reached. So 3 jumps are made.

Input:  arr[] = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}
Output: 10
Explanation: In every step a jump is needed so the count of jumps is 10.

Recommended Practice

## Minimum number of jumps to reach the end using Recursion:

Start from the first element and recursively call for all the elements reachable from the first element. The minimum number of jumps to reach end from first can be calculated using the minimum value from the recursive calls.

minJumps(start, end) = Min ( minJumps(k, end) ) for all k reachable from start.

Follow the steps mentioned below to implement the idea:

• Create a recursive function.
• In each recursive call get all the reachable nodes from that index.
• For each of the index call the recursive function.
• Find the minimum number of jumps to reach the end from current index.
• Return the minimum number of jumps from the recursive call.

Below is the Implementation of the above approach:

## C++

 `// C++ program to find Minimum``// number of jumps to reach end``#include ``using` `namespace` `std;` `// Function to return the minimum number``// of jumps to reach arr[h] from arr[l]``int` `minJumps(``int` `arr[], ``int` `n)``{` `    ``// Base case: when source and``    ``// destination are same``    ``if` `(n == 1)``        ``return` `0;` `    ``// Traverse through all the points``    ``// reachable from arr[l]``    ``// Recursively, get the minimum number``    ``// of jumps needed to reach arr[h] from``    ``// these reachable points``    ``int` `res = INT_MAX;``    ``for` `(``int` `i = n - 2; i >= 0; i--) {``        ``if` `(i + arr[i] >= n - 1) {``            ``int` `sub_res = minJumps(arr, i + 1);``            ``if` `(sub_res != INT_MAX)``                ``res = min(res, sub_res + 1);``        ``}``    ``}` `    ``return` `res;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``cout << ``"Minimum number of jumps to"``;``    ``cout << ``" reach the end is "` `<< minJumps(arr, n);``    ``return` `0;``}` `// This code is contributed``// by Shivi_Aggarwal`

## C

 `// C program to find Minimum``// number of jumps to reach end``#include ``#include ` `// Returns minimum number of``// jumps to reach arr[h] from arr[l]``int` `minJumps(``int` `arr[], ``int` `l, ``int` `h)``{``    ``// Base case: when source and destination are same``    ``if` `(h == l)``        ``return` `0;` `    ``// When nothing is reachable from the given source``    ``if` `(arr[l] == 0)``        ``return` `INT_MAX;` `    ``// Traverse through all the points``    ``// reachable from arr[l]. Recursively``    ``// get the minimum number of jumps``    ``// needed to reach arr[h] from these``    ``// reachable points.``    ``int` `min = INT_MAX;``    ``for` `(``int` `i = l + 1; i <= h && i <= l + arr[l]; i++) {``        ``int` `jumps = minJumps(arr, i, h);``        ``if` `(jumps != INT_MAX && jumps + 1 < min)``            ``min = jumps + 1;``    ``}` `    ``return` `min;``}` `// Driver program to test above function``int` `main()``{``    ``int` `arr[] = { 1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``printf``(``"Minimum number of jumps to reach end is %d "``,``           ``minJumps(arr, 0, n - 1));``    ``return` `0;``}`

## Java

 `// Java program to find Minimum``// number of jumps to reach end``import` `java.io.*;``import` `java.util.*;` `class` `GFG {``    ``// Returns minimum number of``    ``// jumps to reach arr[h] from arr[l]``    ``static` `int` `minJumps(``int` `arr[], ``int` `l, ``int` `h)``    ``{``        ``// Base case: when source``        ``// and destination are same``        ``if` `(h == l)``            ``return` `0``;` `        ``// When nothing is reachable``        ``// from the given source``        ``if` `(arr[l] == ``0``)``            ``return` `Integer.MAX_VALUE;` `        ``// Traverse through all the points``        ``// reachable from arr[l]. Recursively``        ``// get the minimum number of jumps``        ``// needed to reach arr[h] from these``        ``// reachable points.``        ``int` `min = Integer.MAX_VALUE;``        ``for` `(``int` `i = l + ``1``; i <= h && i <= l + arr[l];``             ``i++) {``            ``int` `jumps = minJumps(arr, i, h);``            ``if` `(jumps != Integer.MAX_VALUE``                ``&& jumps + ``1` `< min)``                ``min = jumps + ``1``;``        ``}``        ``return` `min;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `arr[] = { ``1``, ``3``, ``5``, ``8``, ``9``, ``2``, ``6``, ``7``, ``6``, ``8``, ``9` `};``        ``int` `n = arr.length;``        ``System.out.print(``            ``"Minimum number of jumps to reach end is "``            ``+ minJumps(arr, ``0``, n - ``1``));``    ``}``}` `// This code is contributed by Sahil_Bansall`

## Python3

 `# Python3 program to find Minimum``# number of jumps to reach end` `# Returns minimum number of jumps``# to reach arr[h] from arr[l]`  `def` `minJumps(arr, l, h):` `    ``# Base case: when source and``    ``# destination are same``    ``if` `(h ``=``=` `l):``        ``return` `0` `    ``# when nothing is reachable``    ``# from the given source``    ``if` `(arr[l] ``=``=` `0``):``        ``return` `float``(``'inf'``)` `    ``# Traverse through all the points``    ``# reachable from arr[l]. Recursively``    ``# get the minimum number of jumps``    ``# needed to reach arr[h] from``    ``# these reachable points.``    ``min` `=` `float``(``'inf'``)``    ``for` `i ``in` `range``(l ``+` `1``, h ``+` `1``):``        ``if` `(i < l ``+` `arr[l] ``+` `1``):``            ``jumps ``=` `minJumps(arr, i, h)``            ``if` `(jumps !``=` `float``(``'inf'``) ``and``                    ``jumps ``+` `1` `< ``min``):``                ``min` `=` `jumps ``+` `1` `    ``return` `min`  `# Driver program to test above function``arr ``=` `[``1``, ``3``, ``5``, ``8``, ``9``, ``2``, ``6``, ``7``, ``6``, ``8``, ``9``]``n ``=` `len``(arr)``print``(``'Minimum number of jumps to reach'``,``      ``'end is'``, minJumps(arr, ``0``, n``-``1``))` `# This code is contributed by Soumen Ghosh`

## C#

 `// C# program to find Minimum``// number of jumps to reach end``using` `System;` `class` `GFG {``    ``// Returns minimum number of``    ``// jumps to reach arr[h] from arr[l]``    ``static` `int` `minJumps(``int``[] arr, ``int` `l, ``int` `h)``    ``{``        ``// Base case: when source``        ``// and destination are same``        ``if` `(h == l)``            ``return` `0;` `        ``// When nothing is reachable``        ``// from the given source``        ``if` `(arr[l] == 0)``            ``return` `int``.MaxValue;` `        ``// Traverse through all the points``        ``// reachable from arr[l]. Recursively``        ``// get the minimum number of jumps``        ``// needed to reach arr[h] from these``        ``// reachable points.``        ``int` `min = ``int``.MaxValue;``        ``for` `(``int` `i = l + 1; i <= h && i <= l + arr[l];``             ``i++) {``            ``int` `jumps = minJumps(arr, i, h);``            ``if` `(jumps != ``int``.MaxValue && jumps + 1 < min)``                ``min = jumps + 1;``        ``}``        ``return` `min;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int``[] arr = { 1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9 };``        ``int` `n = arr.Length;``        ``Console.Write(``            ``"Minimum number of jumps to reach end is "``            ``+ minJumps(arr, 0, n - 1));``    ``}``}` `// This code is contributed by Sam007`

## PHP

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## Javascript

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Output

`Minimum number of jumps to reach the end is 3`

Time complexity: O(nNn).

• There are maximum n possible ways to move from an element.
• So the maximum number of steps can be nn, Thus O(nn)

Auxiliary Space: O(n). For recursion call stack.

## Minimum number of jumps to reach end using Dynamic Programming from left to right:

It can be observed that there will be overlapping subproblems.

For example in array, arr[] = {1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9} minJumps(3, 9) will be called two times as arr[3] is reachable from arr[1] and arr[2]. So this problem has both properties (optimal substructure and overlapping subproblems) of Dynamic Programming

Follow the below steps to implement the idea:

• Create jumps[] array from left to right such that jumps[i] indicate the minimum number of jumps needed to reach arr[i] from arr[0].
• To fill the jumps array run a nested loop inner loop counter is j and the outer loop count is i.
• Outer loop from 1 to n-1 and inner loop from 0 to i.
• If i is less than j + arr[j] then set jumps[i] to minimum of jumps[i] and jumps[j] + 1. initially set jump[i] to INT MAX
• Return jumps[n-1].

Below is the implementation of the above approach:

## C++

 `// C++ program for Minimum number``// of jumps to reach end``#include ``using` `namespace` `std;` `int` `min(``int` `x, ``int` `y) { ``return` `(x < y) ? x : y; }` `// Returns minimum number of jumps``// to reach arr[n-1] from arr[0]``int` `minJumps(``int` `arr[], ``int` `n)``{``    ``// jumps[n-1] will hold the result``    ``int``* jumps = ``new` `int``[n];``    ``int` `i, j;` `    ``if` `(n == 0 || arr[0] == 0)``        ``return` `INT_MAX;` `    ``jumps[0] = 0;` `    ``// Find the minimum number of jumps to reach arr[i]``    ``// from arr[0], and assign this value to jumps[i]``    ``for` `(i = 1; i < n; i++) {``        ``jumps[i] = INT_MAX;``        ``for` `(j = 0; j < i; j++) {``            ``if` `(i <= j + arr[j] && jumps[j] != INT_MAX) {``                ``jumps[i] = min(jumps[i], jumps[j] + 1);``                ``break``;``            ``}``        ``}``    ``}``    ``return` `jumps[n - 1];``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9 };``    ``int` `size = ``sizeof``(arr) / ``sizeof``(``int``);``    ``cout << ``"Minimum number of jumps to reach end is "``         ``<< minJumps(arr, size);``    ``return` `0;``}` `// This is code is contributed by rathbhupendra`

## C

 `// C program for Minimum number``// of jumps to reach end``#include ``#include ` `int` `min(``int` `x, ``int` `y) { ``return` `(x < y) ? x : y; }` `// Returns minimum number of``// jumps to reach arr[n-1] from arr[0]``int` `minJumps(``int` `arr[], ``int` `n)``{``    ``// jumps[n-1] will hold the result``    ``int` `jumps[n];``    ``int` `i, j;` `    ``if` `(n == 0 || arr[0] == 0)``        ``return` `INT_MAX;` `    ``jumps[0] = 0;` `    ``// Find the minimum number of``    ``// jumps to reach arr[i]``    ``// from arr[0], and assign this``    ``// value to jumps[i]``    ``for` `(i = 1; i < n; i++) {``        ``jumps[i] = INT_MAX;``        ``for` `(j = 0; j < i; j++) {``            ``if` `(i <= j + arr[j] && jumps[j] != INT_MAX) {``                ``jumps[i] = min(jumps[i], jumps[j] + 1);``                ``break``;``            ``}``        ``}``    ``}``    ``return` `jumps[n - 1];``}` `// Driver program to test above function``int` `main()``{``    ``int` `arr[] = { 1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9 };``    ``int` `size = ``sizeof``(arr) / ``sizeof``(``int``);``    ``printf``(``"Minimum number of jumps to reach end is %d "``,``           ``minJumps(arr, size));``    ``return` `0;``}`

## Java

 `// JAVA Code for Minimum number``// of jumps to reach end` `import` `java.io.*;` `class` `GFG {` `    ``private` `static` `int` `minJumps(``int``[] arr, ``int` `n)``    ``{``        ``// jumps[n-1] will hold the``        ``int` `jumps[] = ``new` `int``[n];``        ``// result``        ``int` `i, j;` `        ``// if first element is 0,``        ``if` `(n == ``0` `|| arr[``0``] == ``0``)``            ``return` `Integer.MAX_VALUE;``        ``// end cannot be reached` `        ``jumps[``0``] = ``0``;` `        ``// Find the minimum number of jumps to reach arr[i]``        ``// from arr[0], and assign this value to jumps[i]``        ``for` `(i = ``1``; i < n; i++) {``            ``jumps[i] = Integer.MAX_VALUE;``            ``for` `(j = ``0``; j < i; j++) {``                ``if` `(i <= j + arr[j]``                    ``&& jumps[j] != Integer.MAX_VALUE) {``                    ``jumps[i]``                        ``= Math.min(jumps[i], jumps[j] + ``1``);``                    ``break``;``                ``}``            ``}``        ``}``        ``return` `jumps[n - ``1``];``    ``}` `    ``// driver program to test above function``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``1``, ``3``, ``5``, ``8``, ``9``, ``2``, ``6``, ``7``, ``6``, ``8``, ``9` `};` `        ``System.out.println(``            ``"Minimum number of jumps to reach end is : "``            ``+ minJumps(arr, arr.length));``    ``}``}` `// This code is contributed by Arnav Kr. Mandal.`

## Python3

 `# Python3 program to find Minimum``# number of jumps to reach end` `# Returns minimum number of jumps``# to reach arr[n-1] from arr[0]`  `def` `minJumps(arr, n):``    ``jumps ``=` `[``0` `for` `i ``in` `range``(n)]` `    ``if` `(n ``=``=` `0``) ``or` `(arr[``0``] ``=``=` `0``):``        ``return` `float``(``'inf'``)` `    ``jumps[``0``] ``=` `0` `    ``# Find the minimum number of``    ``# jumps to reach arr[i] from``    ``# arr[0] and assign this``    ``# value to jumps[i]``    ``for` `i ``in` `range``(``1``, n):``        ``jumps[i] ``=` `float``(``'inf'``)``        ``for` `j ``in` `range``(i):``            ``if` `(i <``=` `j ``+` `arr[j]) ``and` `(jumps[j] !``=` `float``(``'inf'``)):``                ``jumps[i] ``=` `min``(jumps[i], jumps[j] ``+` `1``)``                ``break``    ``return` `jumps[n``-``1``]`  `# Driver Program to test above function``arr ``=` `[``1``, ``3``, ``5``, ``8``, ``9``, ``2``, ``6``, ``7``, ``6``, ``8``, ``9``]``size ``=` `len``(arr)``print``(``'Minimum number of jumps to reach'``,``      ``'end is'``, minJumps(arr, size))` `# This code is contributed by Soumen Ghosh`

## C#

 `// C# Code for Minimum number of jumps to reach end``using` `System;` `class` `GFG {``    ``static` `int` `minJumps(``int``[] arr, ``int` `n)``    ``{``        ``// jumps[n-1] will hold the``        ``// result``        ``int``[] jumps = ``new` `int``[n];` `        ``// if first element is 0,``        ``if` `(n == 0 || arr[0] == 0)` `            ``// end cannot be reached``            ``return` `int``.MaxValue;` `        ``jumps[0] = 0;` `        ``// Find the minimum number of``        ``// jumps to reach arr[i]``        ``// from arr[0], and assign``        ``// this value to jumps[i]``        ``for` `(``int` `i = 1; i < n; i++) {``            ``jumps[i] = ``int``.MaxValue;``            ``for` `(``int` `j = 0; j < i; j++) {``                ``if` `(i <= j + arr[j]``                    ``&& jumps[j] != ``int``.MaxValue) {``                    ``jumps[i]``                        ``= Math.Min(jumps[i], jumps[j] + 1);``                    ``break``;``                ``}``            ``}``        ``}``        ``return` `jumps[n - 1];``    ``}` `    ``// Driver program``    ``public` `static` `void` `Main()``    ``{``        ``int``[] arr = { 1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9 };``        ``Console.Write(``            ``"Minimum number of jumps to reach end is : "``            ``+ minJumps(arr, arr.Length));``    ``}``}` `// This code is contributed by Sam007`

## PHP

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## Javascript

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Output

`Minimum number of jumps to reach end is 3`

Thanks to paras for suggesting this method.

Time Complexity: O(n2
Auxiliary Space: O(n), since n extra space has been taken.

Another implementation using Dynamic programming:

Build jumps[] array from right to left such that jumps[i] indicate the minimum number of jumps needed to reach arr[n-1] from arr[i]. Finally, we return jumps[0]. Use Dynamic programming in a similar way of the above method.

Below is the Implementation of the above approach:

## C++

 `// C++ program to find Minimum number of jumps to reach end``#include ``using` `namespace` `std;` `// Returns Minimum number of jumps to reach end``int` `minJumps(``int` `arr[], ``int` `n)``{``    ``// jumps[0] will hold the result``    ``int``* jumps = ``new` `int``[n];``    ``int` `min;` `    ``// Minimum number of jumps needed to reach last element``    ``// from last elements itself is always 0``    ``jumps[n - 1] = 0;` `    ``// Start from the second element, move from right to``    ``// left and construct the jumps[] array where jumps[i]``    ``// represents minimum number of jumps needed to reach``    ``// arr[m-1] from arr[i]``    ``for` `(``int` `i = n - 2; i >= 0; i--) {``        ``// If arr[i] is 0 then arr[n-1] can't be reached``        ``// from here``        ``if` `(arr[i] == 0)``            ``jumps[i] = INT_MAX;` `        ``// If we can directly reach to the end point from``        ``// here then jumps[i] is 1``        ``else` `if` `(arr[i] >= n - i - 1)``            ``jumps[i] = 1;` `        ``// Otherwise, to find out the minimum number of``        ``// jumps needed to reach arr[n-1], check all the``        ``// points reachable from here and jumps[] value for``        ``// those points``        ``else` `{``            ``// initialize min value``            ``min = INT_MAX;` `            ``// following loop checks with all reachable``            ``// points and takes the minimum``            ``for` `(``int` `j = i + 1; j < n && j <= arr[i] + i;``                 ``j++) {``                ``if` `(min > jumps[j])``                    ``min = jumps[j];``            ``}` `            ``// Handle overflow``            ``if` `(min != INT_MAX)``                ``jumps[i] = min + 1;``            ``else``                ``jumps[i] = min; ``// or INT_MAX``        ``}``    ``}` `    ``return` `jumps[0];``}` `// Driver program to test above function``int` `main()``{``    ``int` `arr[] = { 1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9 };``    ``int` `size = ``sizeof``(arr) / ``sizeof``(``int``);``    ``cout << ``"Minimum number of jumps to reach"``         ``<< ``" end is "` `<< minJumps(arr, size);``    ``return` `0;``}` `// This code is contributed by Sania Kumari Gupta`

## C

 `// C program to find Minimum number of jumps to reach end``#include ``#include ` `// Returns Minimum number of jumps to reach end``int` `minJumps(``int` `arr[], ``int` `n)``{``    ``// jumps[0] will hold the result``    ``int` `jumps[n];``    ``int` `min;` `    ``// Minimum number of jumps needed to reach last element``    ``// from last elements itself is always 0``    ``jumps[n - 1] = 0;` `    ``// Start from the second element, move from right to``    ``// left and construct the jumps[] array where jumps[i]``    ``// represents minimum number of jumps needed to reach``    ``// arr[m-1] from arr[i]``    ``for` `(``int` `i = n - 2; i >= 0; i--) {``        ``// If arr[i] is 0 then arr[n-1] can't be reached``        ``// from here``        ``if` `(arr[i] == 0)``            ``jumps[i] = INT_MAX;` `        ``// If we can directly reach to the end point from``        ``// here then jumps[i] is 1``        ``else` `if` `(arr[i] >= n - i - 1)``            ``jumps[i] = 1;` `        ``// Otherwise, to find out the minimum number of``        ``// jumps needed to reach arr[n-1], check all the``        ``// points reachable from here and jumps[] value for``        ``// those points``        ``else` `{``            ``// initialize min value``            ``min = INT_MAX;` `            ``// following loop checks with all reachable``            ``// points and takes the minimum``            ``for` `(``int` `j = i + 1; j < n && j <= arr[i] + i;``                 ``j++) {``                ``if` `(min > jumps[j])``                    ``min = jumps[j];``            ``}` `            ``// Handle overflow``            ``if` `(min != INT_MAX)``                ``jumps[i] = min + 1;``            ``else``                ``jumps[i] = min; ``// or INT_MAX``        ``}``    ``}` `    ``return` `jumps[0];``}` `// Driver program to test above function``int` `main()``{``    ``int` `arr[] = { 1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9 };``    ``int` `size = ``sizeof``(arr) / ``sizeof``(``int``);``    ``printf``(``"Minimum number of jumps to reach end is %d "``,``           ``minJumps(arr, size));``    ``return` `0;``}` `// This code is contributed by Sania Kumari Gupta`

## Java

 `// Java program to find Minimum number of jumps to reach end` `class` `GFG {``    ``// Returns Minimum number of jumps to reach end``    ``static` `int` `minJumps(``int` `arr[], ``int` `n)``    ``{``        ``// jumps[0] will hold the result``        ``int``[] jumps = ``new` `int``[n];``        ``int` `min;` `        ``// Minimum number of jumps needed to reach last``        ``// element from last elements itself is always 0``        ``jumps[n - ``1``] = ``0``;` `        ``// Start from the second element, move from right to``        ``// left and construct the jumps[] array where``        ``// jumps[i] represents minimum number of jumps``        ``// needed to reach arr[m-1] from arr[i]``        ``for` `(``int` `i = n - ``2``; i >= ``0``; i--) {``            ``// If arr[i] is 0 then arr[n-1] can't be reached``            ``// from here``            ``if` `(arr[i] == ``0``)``                ``jumps[i] = Integer.MAX_VALUE;` `            ``// If we can directly reach to the end point``            ``// from here then jumps[i] is 1``            ``else` `if` `(arr[i] >= n - i - ``1``)``                ``jumps[i] = ``1``;` `            ``// Otherwise, to find out the minimum number of``            ``// jumps needed to reach arr[n-1], check all the``            ``// points reachable from here and jumps[] value``            ``// for those points``            ``else` `{``                ``// initialize min value``                ``min = Integer.MAX_VALUE;` `                ``// following loop checks with all reachable``                ``// points and takes the minimum``                ``for` `(``int` `j = i + ``1``;``                     ``j < n && j <= arr[i] + i; j++) {``                    ``if` `(min > jumps[j])``                        ``min = jumps[j];``                ``}` `                ``// Handle overflow``                ``if` `(min != Integer.MAX_VALUE)``                    ``jumps[i] = min + ``1``;``                ``else``                    ``jumps[i] = min; ``// or Integer.MAX_VALUE``            ``}``        ``}` `        ``return` `jumps[``0``];``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int``[] arr = { ``1``, ``3``, ``5``, ``8``, ``9``, ``2``, ``6``, ``7``, ``6``, ``8``, ``9` `};``        ``int` `size = arr.length;``        ``System.out.println(``            ``"Minimum number of jumps to reach end is "``            ``+ minJumps(arr, size));``    ``}``}` `// This code is contributed by Sania Kumari Gupta`

## Python3

 `# Python3 program to find Minimum``# number of jumps to reach end` `# Returns Minimum number of``# jumps to reach end`  `def` `minJumps(arr, n):` `    ``# jumps[0] will hold the result``    ``jumps ``=` `[``0` `for` `i ``in` `range``(n)]` `    ``# Minimum number of jumps needed``    ``# to reach last element from``    ``# last elements itself is always 0``    ``# jumps[n-1] is also initialized to 0` `    ``# Start from the second element,``    ``# move from right to left and``    ``# construct the jumps[] array where``    ``# jumps[i] represents minimum number``    ``# of jumps needed to reach arr[m-1]``    ``# form arr[i]``    ``for` `i ``in` `range``(n``-``2``, ``-``1``, ``-``1``):` `        ``# If arr[i] is 0 then arr[n-1]``        ``# can't be reached from here``        ``if` `(arr[i] ``=``=` `0``):``            ``jumps[i] ``=` `float``(``'inf'``)` `        ``# If we can directly reach to``        ``# the end point from here then``        ``# jumps[i] is 1``        ``elif` `(arr[i] >``=` `n ``-` `i ``-` `1``):``            ``jumps[i] ``=` `1` `        ``# Otherwise, to find out the``        ``# minimum number of jumps``        ``# needed to reach arr[n-1],``        ``# check all the points``        ``# reachable from here and``        ``# jumps[] value for those points``        ``else``:``            ``# initialize min value``            ``min` `=` `float``(``'inf'``)` `            ``# following loop checks with``            ``# all reachable points and``            ``# takes the minimum``            ``for` `j ``in` `range``(i ``+` `1``, n):``                ``if` `(j <``=` `arr[i] ``+` `i):``                    ``if` `(``min` `> jumps[j]):``                        ``min` `=` `jumps[j]` `            ``# Handle overflow``            ``if` `(``min` `!``=` `float``(``'inf'``)):``                ``jumps[i] ``=` `min` `+` `1``            ``else``:``                ``# or INT_MAX``                ``jumps[i] ``=` `min` `    ``return` `jumps[``0``]`  `# Driver program to test above function``arr ``=` `[``1``, ``3``, ``5``, ``8``, ``9``, ``2``, ``6``, ``7``, ``6``, ``8``, ``9``]``n ``=` `len``(arr)``print``(``'Minimum number of jumps to reach'``,``      ``'end is'``, minJumps(arr, n``-``1``))` `# This code is contributed by Soumen Ghosh`

## C#

 `// C# program to find Minimum``// number of jumps to reach end``using` `System;` `class` `GFG {``    ``// Returns Minimum number``    ``// of jumps to reach end``    ``public` `static` `int` `minJumps(``int``[] arr, ``int` `n)``    ``{``        ``// jumps[0] will``        ``// hold the result``        ``int``[] jumps = ``new` `int``[n];``        ``int` `min;` `        ``// Minimum number of jumps needed to``        ``// reach last element from last elements``        ``// itself is always 0``        ``jumps[n - 1] = 0;` `        ``// Start from the second element, move``        ``// from right to left and construct the``        ``// jumps[] array where jumps[i] represents``        ``// minimum number of jumps needed to reach``        ``// arr[m-1] from arr[i]``        ``for` `(``int` `i = n - 2; i >= 0; i--) {``            ``// If arr[i] is 0 then arr[n-1]``            ``// can't be reached from here``            ``if` `(arr[i] == 0) {``                ``jumps[i] = ``int``.MaxValue;``            ``}` `            ``// If we can directly reach to the end``            ``// point from here then jumps[i] is 1``            ``else` `if` `(arr[i] >= n - i - 1) {``                ``jumps[i] = 1;``            ``}` `            ``// Otherwise, to find out the minimum``            ``// number of jumps needed to reach``            ``// arr[n-1], check all the points``            ``// reachable from here and jumps[] value``            ``// for those points``            ``else` `{``                ``// initialize min value``                ``min = ``int``.MaxValue;` `                ``// following loop checks with all``                ``// reachable points and takes the minimum``                ``for` `(``int` `j = i + 1;``                     ``j < n && j <= arr[i] + i; j++) {``                    ``if` `(min > jumps[j]) {``                        ``min = jumps[j];``                    ``}``                ``}` `                ``// Handle overflow``                ``if` `(min != ``int``.MaxValue) {``                    ``jumps[i] = min + 1;``                ``}``                ``else` `{``                    ``jumps[i] = min; ``// or Integer.MAX_VALUE``                ``}``            ``}``        ``}` `        ``return` `jumps[0];``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``int``[] arr``            ``= ``new` `int``[] { 1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9 };``        ``int` `size = arr.Length;``        ``Console.WriteLine(``"Minimum number of"``                          ``+ ``" jumps to reach end is "``                          ``+ minJumps(arr, size));``    ``}``}` `// This code is contributed by Shrikant13`

## PHP

 `= 0; ``\$i``--)``    ``{``        ``// If arr[i] is 0 then arr[n-1]``        ``// can't be reached from here``        ``if` `(``\$arr``[``\$i``] == 0)``            ``\$jumps``[``\$i``] = PHP_INT_MAX;` `        ``// If we can directly reach to``        ``// the end point from here then``        ``// jumps[i] is 1``        ``else` `if` `(``\$arr``[``\$i``] >= (``\$n` `- ``\$i``) - 1)``            ``\$jumps``[``\$i``] = 1;` `        ``// Otherwise, to find out the minimum``        ``// number of jumps needed to reach``        ``// arr[n-1], check all the points``        ``// reachable from here and jumps[]``        ``// value for those points``        ``else``        ``{``            ``// initialize min value``            ``\$min` `= PHP_INT_MAX;` `            ``// following loop checks with all``            ``// reachable points and takes``            ``// the minimum``            ``for` `(``\$j` `= ``\$i` `+ 1; ``\$j` `< ``\$n` `&&``                 ``\$j` `<= ``\$arr``[``\$i``] + ``\$i``; ``\$j``++)``            ``{``                ``if` `(``\$min` `> ``\$jumps``[``\$j``])``                    ``\$min` `= ``\$jumps``[``\$j``];``            ``}` `            ``// Handle overflow``            ``if` `(``\$min` `!= PHP_INT_MAX)``                ``\$jumps``[``\$i``] = ``\$min` `+ 1;``            ``else``                ``\$jumps``[``\$i``] = ``\$min``; ``// or INT_MAX``        ``}``    ``}` `    ``return` `\$jumps``[0];``}` `// Driver Code``\$arr` `= ``array``(1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9);``\$size` `= sizeof(``\$arr``);``echo` `"Minimum number of jumps to reach"``,``     ``" end is "``, minJumps(``\$arr``, ``\$size``);` `// This code is contributed by ajit.``?>`

## Javascript

 ``

Output

`Minimum number of jumps to reach end is 3`

Time complexity: O(n2). Nested traversal of the array is needed.
Auxiliary Space: O(n). To store the DP array linear space is needed.

Minimum number of jumps to reach end | Set 2 (O(n) solution)
Thanks to Ashish for suggesting this solution.