Minimum number of jumps to reach end | Set 2 (O(n) solution)
Given an array of integers where each element represents the max number of steps that can be made forward from that element. Write a function to return the minimum number of jumps to reach the end of the array (starting from the first element). If an element is 0, then we cannot move through that element.
Examples:
Input: arr[] = {1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9}
Output: 3 (1-> 3 -> 8 -> 9)
Explanation: Jump from 1st element to
2nd element as there is only 1 step,
now there are three options 5, 8 or 9.
If 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.
In this post, its O(n) solution will be discussed.
In Set -1, O(n2) solution is discussed.
Implementation:
Variables to be used:
- maxReach The variable maxReach stores at all time the maximal reachable index in the array.
- step The variable step stores the number of steps we can still take(and is initialized with value at index 0, i.e. initial number of steps)
- jump jump stores the amount of jumps necessary to reach that maximal reachable position.
Given array arr = 1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9
- maxReach = arr[0]; // arr[0] = 1, so the maximum index we can reach at the moment is 1.
step = arr[0]; // arr[0] = 1, the amount of steps we can still take is also 1.
jump = 1; // we will always need to take at least one jump. - Now, starting iteration from index 1, the above values are updated as follows:
-
First we test whether we have reached the end of the array, in that case we just need to return the jump variable.
if (i == arr.length - 1) return jump; - Next we update the maxReach. This is equal to the maximum of maxReach and i+arr[i](the number of steps we can take from the current position).
maxReach = Math.max(maxReach, i+arr[i]);
- We used up a step to get to the current index, so steps has to be decreased.
step--;
- If no more steps are remaining (i.e. steps=0, then we must have used a jump. Therefore increase jump. Since we know that it is possible somehow to reach maxReach, we again initialize the steps to the number of steps to reach maxReach from position i. But before re-initializing step, we also check whether a step is becoming zero or negative. In this case, It is not possible to reach further.
if (step == 0) { jump++; if(i>=maxReach) return -1; step = maxReach - i; }
-
First we test whether we have reached the end of the array, in that case we just need to return the jump variable.
C++
// C++ program to count Minimum number // of jumps to reach end #include <bits/stdc++.h> using namespace std; int max(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) { // The number of jumps needed to // reach the starting index is 0 if (n <= 1) return 0; // Return -1 if not possible to jump if (arr[0] == 0) return -1; // initialization // stores all time the maximal // reachable index in the array. int maxReach = arr[0]; // stores the number of steps // we can still take int step = arr[0]; // stores the number of jumps // necessary to reach that maximal // reachable position. int jump = 1; // Start traversing array int i = 1; for (i = 1; i < n; i++) { // Check if we have reached the end of the array if (i == n - 1) return jump; // updating maxReach maxReach = max(maxReach, i + arr[i]); // we use a step to get to the current index step--; // If no further steps left if (step == 0) { // we must have used a jump jump++; // Check if the current index/position or lesser index // is the maximum reach point from the previous indexes if (i >= maxReach) return -1; // re-initialize the steps to the amount // of steps to reach maxReach from position i. step = maxReach - i; } } return -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); // Calling the minJumps function cout << ("Minimum number of jumps to reach end is %d ", minJumps(arr, size)); return 0; } // This code is contributed by // Shashank_Sharma |
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C
// C program to count Minimum number // of jumps to reach end #include <stdio.h> int max(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) { // The number of jumps needed to // reach the starting index is 0 if (n <= 1) return 0; // Return -1 if not possible to jump if (arr[0] == 0) return -1; // initialization // stores all time the maximal // reachable index in the array. int maxReach = arr[0]; // stores the number of steps // we can still take int step = arr[0]; // stores the number of jumps // necessary to reach that maximal // reachable position. int jump = 1; // Start traversing array int i = 1; for (i = 1; i < n; i++) { // Check if we have reached the end of the array if (i == n - 1) return jump; // updating maxReach maxReach = max(maxReach, i + arr[i]); // we use a step to get to the current index step--; // If no further steps left if (step == 0) { // we must have used a jump jump++; // Check if the current index/position or lesser index // is the maximum reach point from the previous indexes if (i >= maxReach) return -1; // re-initialize the steps to the amount // of steps to reach maxReach from position i. step = maxReach - i; } } return -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); // Calling the minJumps function printf( "Minimum number of jumps to reach end is %d ", minJumps(arr, size)); return 0; } // This code is contributed by Abhishek Kumar Singh |
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Java
// Java program to count Minimum number // of jumps to reach end class Test { static int minJumps(int arr[]) { if (arr.length <= 1) return 0; // Return -1 if not possible to jump if (arr[0] == 0) return -1; // initialization int maxReach = arr[0]; int step = arr[0]; int jump = 1; // Start traversing array for (int i = 1; i < arr.length; i++) { // Check if we have reached // the end of the array if (i == arr.length - 1) return jump; // updating maxReach maxReach = Math.max(maxReach, i + arr[i]); // we use a step to get to the current index step--; // If no further steps left if (step == 0) { // we must have used a jump jump++; // Check if the current // index/position or lesser index // is the maximum reach point // from the previous indexes if (i >= maxReach) return -1; // re-initialize the steps to the amount // of steps to reach maxReach from position i. step = maxReach - i; } } return -1; } // Driver method to test the above function public static void main(String[] args) { int arr[] = new int[] { 1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9 }; // calling minJumps method System.out.println(minJumps(arr)); } } |
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Python
# python program to count Minimum number # of jumps to reach end # Returns minimum number of jumps to reach arr[n-1] from arr[0] def minJumps(arr, n): # The number of jumps needed to reach the starting index is 0 if (n <= 1): return 0 # Return -1 if not possible to jump if (arr[0] == 0): return -1 # initialization # stores all time the maximal reachable index in the array maxReach = arr[0] # stores the amount of steps we can still take step = arr[0] # stores the amount of jumps necessary to reach that maximal reachable position jump = 1 # Start traversing array for i in range(1, n): # Check if we have reached the end of the array if (i == n-1): return jump # updating maxReach maxReach = max(maxReach, i + arr[i]) # we use a step to get to the current index step -= 1; # If no further steps left if (step == 0): # we must have used a jump jump += 1 # Check if the current index / position or lesser index # is the maximum reach point from the previous indexes if(i >= maxReach): return -1 # re-initialize the steps to the amount # of steps to reach maxReach from position i. step = maxReach - i; return -1 # Driver program to test above function arr = [1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9] size = len(arr) # Calling the minJumps function print("Minimum number of jumps to reach end is % d " % minJumps(arr, size)) # This code is contributed by Aditi Sharma |
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C#
// C# program to count Minimum // number of jumps to reach end using System; class GFG { static int minJumps(int[] arr) { if (arr.Length <= 1) return 0; // Return -1 if not // possible to jump if (arr[0] == 0) return -1; // initialization int maxReach = arr[0]; int step = arr[0]; int jump = 1; // Start traversing array for (int i = 1; i < arr.Length; i++) { // Check if we have reached // the end of the array if (i == arr.Length - 1) return jump; // updating maxReach maxReach = Math.Max(maxReach, i + arr[i]); // we use a step to get // to the current index step--; // If no further steps left if (step == 0) { // we must have used a jump jump++; // Check if the current index/position // or lesser index is the maximum reach // point from the previous indexes if (i >= maxReach) return -1; // re-initialize the steps to // the amount of steps to reach // maxReach from position i. step = maxReach - i; } } return -1; } // Driver Code public static void Main() { int[] arr = new int[] { 1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9 }; // calling minJumps method Console.Write(minJumps(arr)); } } // This code is contributed // by nitin mittal |
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PHP
<?php // PHP program to count Minimum number // of jumps to reach end // Returns minimum number of jumps to reach arr[n-1] from arr[0] function minJumps(&$arr, $n) { // The number of jumps needed to reach the starting index is 0 if ($n <= 1) return 0; // Return -1 if not possible to jump if ($arr[0] == 0) return -1; // initialization // stores all time the maximal reachable index in the array. $maxReach = $arr[0]; // stores the number of steps we can still take $step = $arr[0]; //stores the number of jumps necessary to reach that // maximal reachable position. $jump =1; // Start traversing array $i=1; for ($i = 1; $i < $n; $i++) { // Check if we have reached the end of the array if ($i == $n-1) return $jump; // updating maxReach $maxReach = max($maxReach, $i+$arr[$i]); // we use a step to get to the current index $step--; // If no further steps left if ($step == 0) { // we must have used a jump $jump++; // Check if the current index/position or lesser index // is the maximum reach point from the previous indexes if($i >= $maxReach) return -1; // re-initialize the steps to the amount // of steps to reach maxReach from position i. $step = $maxReach - $i; } } return -1; } // Driver program to test above function $arr=array(1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9); $size = sizeof($arr)/sizeof($arr[0]); // Calling the minJumps function echo "Minimum number of jumps to reach end is " . minJumps($arr, $size); return 0; // This code is contribute by Ita_c. ?> |
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Output:
3
Complexity Analysis:
- Time complexity: O(n).
Only one traversal of the array is needed. - Auxiliary Space: O(1).
There is no space required.
References: Stackoverflow
Thanks to Chiranjeev Jain for suggesting this solution.
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