Minimum jumps to same value or adjacent to reach end of Array

Last Updated : 02 Dec, 2022

Given an array arr[] of size N, the task is to find the minimum number of jumps to reach the last index of the array starting from index 0. In one jump you can move from current index i to index j, if arr[i] = arr[j] and i != j or you can jump to (i + 1) or (i â€“ 1).

Note: You can not jump outside of the array at any time.

Examples:

Input: arr = {100, -23, -23, 404, 100, 23, 23, 23, 3, 404}
Output: 3
Explanation: Valid jump indices are 0 -> 4 -> 3 -> 9.

Input: arr = {7, 6, 9, 6, 9, 6, 9, 7}
Output: 1

An approach using BFS:

Here consider the elements which are at (i + 1), (i – 1), and all elements similar to arr[i] and insert them into a queue to perform BFS. Repeat the BFS in this manner and keep the track of level. When the end of array is reached return the level value.

Follow the steps below to implement the above idea:

• Initialize a map for mapping elements with the indices of their occurrence.
• Initialize a queue and an array visited[] to keep track of the elements that are visited.
• Push starting element into the queue and mark it as visited
• Initialize a variable count for counting the minimum number of valid jumps to reach the last index
• Do the following while the queue size is greater than 0:
• Iterate on all the elements of the queue
• Fetch the front element and pop out from the queue
• Check if we reach the last index or not
• If true, then return the count
• Check if curr + 1 is a valid position to visit or not
• If true, push curr + 1 into the queue and mark it as visited
• Check if curr – 1 is a valid position to visit or not
• If true, push curr – 1 into the queue and mark it as visited
• Now, Iterate over all the elements that are similar to curr
• Check if the child is in a valid position to visit or not
• If true, push the child into the queue and mark it as visited
• Erase all the occurrences of curr from the map because we already considered these elements for a valid jump in the above step
• Increment the count of jump
• Finally, return the count.

Below is the implementation of the above approach.

C++

 `// C++ code to implement the above approach`   `#include ` `using` `namespace` `std;`   `// Function to find the` `// minimum number of jumps required` `int` `minimizeJumps(vector<``int``>& arr)` `{` `    ``int` `n = arr.size();`   `    ``// Initialize a map for mapping element` `    ``// with indices of all similar value` `    ``// occurrences in array` `    ``unordered_map<``int``, vector<``int``> > unmap;`   `    ``// Mapping element with indices` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``unmap[arr[i]].push_back(i);` `    ``}`   `    ``queue<``int``> q;` `    ``vector<``bool``> visited(n, ``false``);`   `    ``// Push starting element into queue` `    ``// and mark it visited` `    ``q.push(0);` `    ``visited[0] = ``true``;`   `    ``// Initialize a variable count for` `    ``// counting the minimum number number` `    ``// of valid jump to reach at last index` `    ``int` `count = 0;`   `    ``// Do while queue size is` `    ``// greater than 0` `    ``while` `(q.size() > 0) {` `        ``int` `size = q.size();`   `        ``// Iterate on all the` `        ``// elements of queue` `        ``for` `(``int` `i = 0; i < size; i++) {`   `            ``// Fetch the front element and` `            ``// pop out from queue` `            ``int` `curr = q.front();` `            ``q.pop();`   `            ``// Check if we reach at the` `            ``// last index or not if true,` `            ``// then return the count` `            ``if` `(curr == n - 1) {` `                ``return` `count;` `            ``}`   `            ``// Check if curr + 1 is valid` `            ``// position to visit or not` `            ``if` `(curr + 1 < n` `                ``&& visited[curr + 1] == ``false``) {`   `                ``// If true, push curr + 1` `                ``// into queue and mark` `                ``// it as visited` `                ``q.push(curr + 1);` `                ``visited[curr + 1] = ``true``;` `            ``}`   `            ``// Check if curr - 1 is valid` `            ``// position to visit or not` `            ``if` `(curr - 1 >= 0` `                ``&& visited[curr - 1] == ``false``) {`   `                ``// If true, push curr - 1` `                ``// into queue and mark` `                ``// it as visited` `                ``q.push(curr - 1);` `                ``visited[curr - 1] = ``true``;` `            ``}`   `            ``// Now, Iterate over all the` `            ``// element that are similar` `            ``// to curr` `            ``for` `(``auto` `child : unmap[arr[curr]]) {` `                ``if` `(curr == child)` `                    ``continue``;`   `                ``// Check if child is valid` `                ``// position to visit or not` `                ``if` `(visited[child] == ``false``) {`   `                    ``// If true, push child` `                    ``// into queue and mark` `                    ``// it as visited` `                    ``q.push(child);` `                    ``visited[child] = ``true``;` `                ``}` `            ``}`   `            ``// Erase all the occurrences` `            ``// of curr from map. Because` `            ``// we already considered these` `            ``// element for valid jump` `            ``// in above step` `            ``unmap.erase(arr[curr]);` `        ``}`   `        ``// Increment the count of jump` `        ``count++;` `    ``}`   `    ``// Finally, return the count.` `    ``return` `count;` `}`   `// Driver code` `int` `main()` `{` `    ``vector<``int``> arr` `        ``= { 100, -23, -23, 404, 100, 23, 23, 23, 3, 404 };`   `    ``// Function Call` `    ``cout << minimizeJumps(arr);`   `    ``return` `0;` `}`

Java

 `// Java code for the above approach` `import` `java.io.*;` `import` `java.util.*;`   `class` `GFG {`   `  ``// Function to find the minimum` `  ``// number of jumps required` `  ``static` `int` `minimizeJumps(``int``[] arr)` `  ``{` `    ``int` `n = arr.length;`   `    ``// Initialize a map for mapping element` `    ``// with indices of all similar value` `    ``// occurrences in array` `    ``HashMap > unmap` `      ``= ``new` `HashMap<>();`   `    ``// Mapping element with indices` `    ``for` `(``int` `i = ``0``; i < n; i++) {` `      ``if` `(unmap.containsKey(arr[i])) {` `        ``unmap.get(arr[i]).add(i);` `      ``}` `      ``else` `{` `        ``List temp = ``new` `ArrayList<>();` `        ``temp.add(i);` `        ``unmap.put(arr[i], temp);` `      ``}` `    ``}`   `    ``Queue q = ``new` `LinkedList<>();` `    ``boolean``[] visited = ``new` `boolean``[n];` `    ``Arrays.fill(visited, ``false``);`   `    ``// Push starting element into queue` `    ``// and mark it visited` `    ``q.add(``0``);` `    ``visited[``0``] = ``true``;`   `    ``// Initialize a variable count for` `    ``// counting the minimum number number` `    ``// of valid jump to reach at last index` `    ``int` `count = ``0``;`   `    ``// Do while queue size is` `    ``// greater than 0` `    ``while` `(q.size() > ``0``) {` `      ``int` `size = q.size();`   `      ``// Iterate on all the` `      ``// elements of queue` `      ``for` `(``int` `i = ``0``; i < size; i++) {`   `        ``// Fetch the front element and` `        ``// pop out from queue` `        ``int` `curr = q.poll();`   `        ``// Check if we reach at the` `        ``// last index or not if true,` `        ``// then return the count` `        ``if` `(curr == n - ``1``) {` `          ``return` `count / ``2``;` `        ``}`   `        ``// Check if curr + 1 is valid` `        ``// position to visit or not` `        ``if` `(curr + ``1` `< n` `            ``&& visited[curr + ``1``] == ``false``) {` `          ``// If true, push curr + 1` `          ``// into queue and mark` `          ``// it as visited` `          ``q.add(curr + ``1``);` `          ``visited[curr + ``1``] = ``true``;` `        ``}`   `        ``// Check if curr - 1 is valid` `        ``// position to visit or not` `        ``if` `(curr - ``1` `>= ``0` `            ``&& visited[curr - ``1``] == ``false``) {`   `          ``// If true, push curr - 1` `          ``// into queue and mark` `          ``// it as visited` `          ``q.add(curr - ``1``);` `          ``visited[curr - ``1``] = ``true``;` `        ``}`   `        ``// Now, Iterate over all the` `        ``// element that are similar` `        ``// to curr` `        ``if` `(unmap.containsKey(arr[i])) {` `          ``for` `(``int` `j = ``0``;` `               ``j < unmap.get(arr[curr]).size();` `               ``j++) {` `            ``int` `child` `              ``= unmap.get(arr[curr]).get(j);` `            ``if` `(curr == child) {` `              ``continue``;` `            ``}` `            ``// Check if child is valid` `            ``// position to visit or not` `            ``if` `(visited[child] == ``false``) {`   `              ``// If true, push child` `              ``// into queue and mark` `              ``// it as visited` `              ``q.add(child);` `              ``visited[child] = ``true``;` `            ``}` `          ``}` `        ``}` `        ``// Erase all the occurrences` `        ``// of curr from map. Because` `        ``// we already considered these` `        ``// element for valid jump` `        ``// in above step` `        ``unmap.remove(arr[curr]);` `      ``}` `      ``// Increment the count of jump` `      ``count++;` `    ``}` `    ``// Finally, return the count.` `    ``return` `count / ``2``;` `  ``}`   `  ``public` `static` `void` `main(String[] args)` `  ``{` `    ``int``[] arr = { ``100``, -``23``, -``23``, ``404``, ``100``,` `                 ``23``,  ``23``,  ``23``,  ``3``,   ``404` `};`   `    ``// Function call` `    ``System.out.print(minimizeJumps(arr));` `  ``}` `}`   `// This code is contributed by lokesh`

Python3

 `# Python code to implement the above approach`   `# Function to find the` `# minimum number of jumps required` `def` `minimizeJumps(arr):` `    ``n ``=` `len``(arr)` `    `  `    ``# Initialize a map for mapping element` `    ``# with indices of all similar value` `    ``# occurrences in array` `    ``unmap ``=` `{}` `    `  `    ``# Mapping element with indices` `    ``for` `i ``in` `range``(n):` `        ``if` `arr[i] ``in` `unmap:` `            ``unmap.get(arr[i]).append(i)` `        ``else``:` `            ``unmap.update({arr[i]:[i]})` `    `  `    ``q ``=` `[]` `    ``visited ``=` `[``False``]``*``n` `    `  `    ``# Push starting element into queue` `    ``# and mark it visited` `    ``q.append(``0``)` `    ``visited[``0``] ``=` `True` `    `  `    ``# Initialize a variable count for` `    ``# counting the minimum number number` `    ``# of valid jump to reach at last index` `    ``count ``=` `0` `    `  `    ``# Do while queue size is` `    ``# greater than 0` `    ``while``(``len``(q) > ``0``):` `        ``size ``=` `len``(q)` `        `  `        ``# Iterate on all the` `        ``# elements of queue` `        ``for` `i ``in` `range``(size):` `            ``# Fetch the front element and` `            ``# pop out from queue` `            ``curr ``=` `q[``0``]` `            ``q.pop(``0``)` `            `  `            ``# Check if we reach at the` `            ``# last index or not if true,` `            ``# then return the count` `            ``if``(curr ``=``=` `n ``-` `1``):` `                ``return` `count``/``/``2` `            `  `            ``# Check if curr + 1 is valid` `            ``# position to visit or not` `            ``if``(curr ``+` `1` `< n ``and` `visited[curr ``+` `1``] ``=``=` `False``):` `                ``# If true, push curr + 1` `                ``# into queue and mark` `                ``# it as visited` `                ``q.append(curr ``+` `1``)` `                ``visited[curr ``+` `1``] ``=` `True` `            `  `            ``# Check if curr - 1 is valid` `            ``# position to visit or not` `            ``if``(curr ``-` `1` `>``=` `0` `and` `visited[curr ``-` `1``] ``=``=` `False``):` `                ``# If true, push curr - 1` `                ``# into queue and mark` `                ``# it as visited` `                ``q.append(curr ``-` `1``)` `                ``visited[curr ``-` `1``] ``=` `True` `            `  `            ``# Now, Iterate over all the` `            ``# element that are similar` `            ``# to curr` `            ``if` `arr[i] ``in` `unmap:` `                ``for` `j ``in` `range``(``len``(unmap[arr[curr]])):` `                    ``child``=``unmap.get(arr[curr])[j]` `                    ``if``(curr``=``=``child):` `                        ``continue` `                    `  `                    ``# Check if child is valid` `                    ``# position to visit or not` `                    ``if``(visited[child] ``=``=` `False``):` `                        ``# If true, push child` `                        ``# into queue and mark` `                        ``# it as visited` `                        ``q.append(child)` `                        ``visited[child] ``=` `True` `            `  `            ``# Erase all the occurrences` `            ``# of curr from map. Because` `            ``# we already considered these` `            ``# element for valid jump` `            ``# in above step` `            ``if` `arr[curr] ``in` `unmap:` `                ``unmap.pop(arr[curr])` `            `  `        `  `        ``# Increment the count of jump` `        ``count ``=` `count ``+` `1` `    `  `    ``# Finally, return the count.` `    ``return` `count``/``/``2`   `# Driver code` `arr``=``[``100``, ``-``23``, ``-``23``, ``404``, ``100``, ``23``, ``23``, ``23``, ``3``, ``404``]`   `# Function Call` `print``(minimizeJumps(arr))`   `# This code is contributed by Pushpesh Raj.`

C#

 `// C# code for the above approach` `using` `System;` `using` `System.Collections.Generic;`   `public` `class` `GFG{` `    `  `    ``// Function to find the minimum` `// number of jumps required` `static` `int` `minimizeJumps(``int``[] arr)` `{` `    ``int` `n = arr.Length;`   `    ``// Initialize a map for mapping element` `    ``// with indices of all similar value` `    ``// occurrences in array` `    ``Dictionary<``int``,List<``int``>> unmap = ``new` `Dictionary<``int``,List<``int``>>();`   `    ``// Mapping element with indices` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``if` `(unmap.ContainsKey(arr[i])) {` `        ``unmap[arr[i]].Add(i);` `      ``}` `      ``else` `{` `        ``List<``int``> temp = ``new` `List<``int``>();` `        ``temp.Add(i);` `        ``unmap.Add(arr[i],temp);` `      ``}` `    ``}`   `    ``List<``int``> q = ``new` `List<``int``>();` `    ``bool``[] visited = ``new` `bool``[n];` `    ``for``(``int` `i=0;i 0) {` `    ``int` `size = q.Count;`   `    ``// Iterate on all the` `    ``// elements of queue` `    ``for` `(``int` `i = 0; i < size; i++) {`   `        ``// Fetch the front element and` `        ``// pop out from queue` `        ``int` `curr = q[0];` `        ``q.RemoveAt(0);`   `        ``// Check if we reach at the` `        ``// last index or not if true,` `        ``// then return the count` `        ``if` `(curr == n - 1) {` `        ``return` `count / 2;` `        ``}`   `        ``// Check if curr + 1 is valid` `        ``// position to visit or not` `        ``if` `(curr + 1 < n` `            ``&& visited[curr + 1] == ``false``) {` `        ``// If true, push curr + 1` `        ``// into queue and mark` `        ``// it as visited` `        ``q.Add(curr + 1);` `        ``visited[curr + 1] = ``true``;` `        ``}`   `        ``// Check if curr - 1 is valid` `        ``// position to visit or not` `        ``if` `(curr - 1 >= 0` `            ``&& visited[curr - 1] == ``false``) {`   `        ``// If true, push curr - 1` `        ``// into queue and mark` `        ``// it as visited` `        ``q.Add(curr - 1);` `        ``visited[curr - 1] = ``true``;` `        ``}`   `        ``// Now, Iterate over all the` `        ``// element that are similar` `        ``// to curr` `        ``if` `(unmap.ContainsKey(arr[i])){` `        ``for` `(``int` `j = 0; j < unmap[arr[curr]].Count; j++) {` `            ``int` `child= unmap[arr[curr]][j];` `            ``if` `(curr == child) {` `            ``continue``;` `            ``}` `            ``// Check if child is valid` `            ``// position to visit or not` `            ``if` `(visited[child] == ``false``) {`   `            ``// If true, push child` `            ``// into queue and mark` `            ``// it as visited` `            ``q.Add(child);` `            ``visited[child] = ``true``;` `            ``}` `        ``}` `        ``}` `        ``// Erase all the occurrences` `        ``// of curr from map. Because` `        ``// we already considered these` `        ``// element for valid jump` `        ``// in above step` `        ``unmap.Remove(arr[curr]);` `    ``}` `    ``// Increment the count of jump` `    ``count++;` `    ``}` `    ``// Finally, return the count.` `    ``return` `count / 2;` `}`   `    ``static` `public` `void` `Main (``string``[] args){` `    ``int``[] arr = { 100, -23, -23, 404, 100,` `                ``23, 23, 23, 3, 404 };`   `    ``// Function call` `    ``Console.WriteLine(minimizeJumps(arr));` `    ``}` `}`     `// This code is contributed by Aman Kumar`

Javascript

 `// JavaScript code for the above approach`   `// Function to find the` `// minimum number of jumps required` `function` `minimizeJumps(arr)` `{` `    ``let n = arr.length;`   `    ``// Initialize a map for mapping element` `    ``// with indices of all similar value` `    ``// occurrences in array` `    ``let unmap = ``new` `Map();`   `    ``// Mapping element with indices` `    ``for` `(let i = 0; i < n; i++) {` `        ``if` `(unmap.has(arr[i]))` `            ``unmap.get(arr[i]).push(i);` `        ``else` `            ``unmap.set(arr[i], [i]);` `    ``}`   `    ``let q = [];` `    ``let visited = ``new` `Array(n).fill(0);`   `    ``// Push starting element into queue` `    ``// and mark it visited` `    ``q.push(0);` `    ``visited[0] = ``true``;`   `    ``// Initialize a variable count for` `    ``// counting the minimum number number` `    ``// of valid jump to reach at last index` `    ``let count = 0;`   `    ``// Do while queue size is` `    ``// greater than 0` `    ``while` `(q.length > 0) {` `        ``let size = q.length;`   `        ``// Iterate on all the` `        ``// elements of queue` `        ``for` `(let i = 0; i < size; i++) {`   `            ``// Fetch the front element and` `            ``// pop out from queue` `            ``let curr = q.shift();`   `            ``// Check if we reach at the` `            ``// last index or not if true,` `            ``// then return the count` `            ``if` `(curr == n - 1) {` `                ``return` `count / 2;` `            ``}`   `            ``// Check if curr + 1 is valid` `            ``// position to visit or not` `            ``if` `(curr + 1 < n` `                ``&& visited[curr + 1] == ``false``) {`   `                ``// If true, push curr + 1` `                ``// into queue and mark` `                ``// it as visited` `                ``q.push(curr + 1);` `                ``visited[curr + 1] = ``true``;` `            ``}`   `            ``// Check if curr - 1 is valid` `            ``// position to visit or not` `            ``if` `(curr - 1 >= 0` `                ``&& visited[curr - 1] == ``false``) {`   `                ``// If true, push curr - 1` `                ``// into queue and mark` `                ``// it as visited` `                ``q.push(curr - 1);` `                ``visited[curr - 1] = ``true``;` `            ``}`   `            ``// Now, Iterate over all the` `            ``// element that are similar` `            ``// to curr` `            ``if` `(unmap.has(arr[i])) {` `                ``for` `(let i = 0;` `                     ``i < unmap.get(arr[curr]).length; i++) {` `                    ``child = unmap.get(arr[curr])[i];` `                    ``if` `(curr == child)` `                        ``continue``;`   `                    ``// Check if child is valid` `                    ``// position to visit or not` `                    ``if` `(visited[child] == ``false``) {`   `                        ``// If true, push child` `                        ``// into queue and mark` `                        ``// it as visited` `                        ``q.push(child);` `                        ``visited[child] = ``true``;` `                    ``}` `                ``}` `            ``}` `            ``// Erase all the occurrences` `            ``// of curr from map. Because` `            ``// we already considered these` `            ``// element for valid jump` `            ``// in above step` `            ``unmap.``delete``(arr[curr]);` `        ``}`   `        ``// Increment the count of jump` `        ``count++;` `    ``}`   `    ``// Finally, return the count.` `    ``return` `count / 2;` `}`   `// Driver code`   `let arr = [ 100, -23, -23, 404, 100, 23, 23, 23, 3, 404 ];`   `// Function Call` `console.log(minimizeJumps(arr));`   `// This code is contributed by Potta Lokesh`

Output

`3`

Time Complexity: O(N)
Auxiliary Space: O(N)