# Minimum number of jumps to obtain an element of opposite parity

Given two arrays **arr[]** and **jumps[]** consisting of **N** positive integers, the task for each array element **arr[i]** is to find the minimum number of jumps required to reach an element of opposite parity. The only possible jumps from any array element **arr[i]** are **(i + jumps[i])** or **(i – jumps[i])**.

**Examples:**

Input:arr[] = {4, 2, 5, 2, 1}, jumps[] = {1, 2, 3, 1, 2}Output:3 2 -1 1 -1Explanation:

Below are the minimum steps required for each element:arr[4]:Since jumps[4] = 2, the only possible jump is (4 – jumps[4]) = 2. Snce arr[2] is of same parity as arr[4] and no further jump within the range of array indice is possible, print-1.arr[3]:Since jumps[3] = 1, both the jumps (3 + jumps[3]) = 4 or (3 – jumps[3]) = 2 are possible. Since both the elements arr[2] and arr[4] are of opposite parity with arr[3], print1.arr[2]:Print-1.arr[1]:Only possible jump is (2 + jumps[2]). Since arr[3] is of same parity as arr[2] and minimum jumps required from arr[3] to reach an array element of opposite parity is 1, the number of jumps required is 2.arr[0]:arr[0] -> arr[3] -> (arr[4] or arr[2]). Therefore, minimum jumps required is 3.

Input:arr[] = {4, 5, 7, 6, 7, 5, 4, 4, 6, 4}, jumps[] = {1, 2, 3, 3, 5, 2, 7, 2, 4, 1}Output:1 1 2 2 1 1 -1 1 1 2

**Naive Approach:** The simplest approach is to solve the problem is to traverse the array and perform Breadth-First Traversal for each array element **arr[i]**, by repeatedly converting to **arr[i – jumps[i]]** and **arr[i + jumps[i]]**, until any invalid index occurs, and after each conversion, check if the array element is of opposite parity to the previous element or not. Print the minimum number of jumps required for each array element accordingly.

**Time Complexity:** O(N^{2})**Auxiliary Space:** O(N)

**Efficient Approach:** To optimize the above approach, the idea is to use multi-source BFS for even and odd array elements individually. Follow the steps below to solve the problem:

- Initialize a vector, say
**ans[]**, to store the minimum jumps required for each array element to reach an element of opposite parity. - Initialize an array of vectors,
**Adj[]**to store the adjacency list of the graph generated. - For every pair
**(i, j)**of valid jumps for each index,**i**of the array create an inverted graph by initializing edges as**(j, i)**. - Perform
**multisource-BFS traversal**for odd elements. Perform the following steps:- Push all indices that contains odd array elements, into the queue and mark all these nodes as visited simultaneously.
- Iterate until queue is non-empty and perform the following operations:
- Pop the node present at the front of the queue and check if any of its now connected to it is of same parity or not. If found to be true, update the
**distance of the child node**as the**1 + distance of parent node.** - Mark the child node visited and push it into the queue.

- Pop the node present at the front of the queue and check if any of its now connected to it is of same parity or not. If found to be true, update the

- Perform
**multisource-BFS traversal**for even elements similarly and store the distances in**ans[]**. - After completing the above steps, print the values stored in
**ans[]**as the result.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Bfs for odd numbers are source` `void` `bfs(` `int` `n, vector<` `int` `>& a,` ` ` `vector<` `int` `> invGr[],` ` ` `vector<` `int` `>& ans, ` `int` `parity)` `{` ` ` `// Initialize queue` ` ` `queue<` `int` `> q;` ` ` `// Stores for each node, the nodes` ` ` `// visited and their distances` ` ` `vector<` `int` `> vis(n + 1, 0);` ` ` `vector<` `int` `> dist(n + 1, 0);` ` ` `// Push odd and even numbers` ` ` `// as sources to the queue` ` ` `// If parity is 0 -> odd` ` ` `// Otherwise -> even` ` ` `for` `(` `int` `i = 1; i <= n; i++) {` ` ` `if` `((a[i] + parity) & 1) {` ` ` `q.push(i);` ` ` `vis[i] = 1;` ` ` `}` ` ` `}` ` ` `// Perform multi-source bfs` ` ` `while` `(!q.empty()) {` ` ` `// Extract the front element` ` ` `// of the queue` ` ` `int` `v = q.front();` ` ` `q.pop();` ` ` `// Traverse nodes connected` ` ` `// to the current node` ` ` `for` `(` `int` `u : invGr[v]) {` ` ` `// If u is not visited` ` ` `if` `(!vis[u]) {` ` ` `dist[u] = dist[v] + 1;` ` ` `vis[u] = 1;` ` ` `// If element with opposite` ` ` `// parity is obtained` ` ` `if` `((a[u] + parity) % 2 == 0) {` ` ` `if` `(ans[u] == -1)` ` ` `// Store its distance from` ` ` `// source in ans[]` ` ` `ans[u] = dist[u];` ` ` `}` ` ` `// Push the current neighbour` ` ` `// to the queue` ` ` `q.push(u);` ` ` `}` ` ` `}` ` ` `}` `}` `// Function to find the minimum jumps` `// required by each index to reach` `// element of opposite parity` `void` `minJumps(vector<` `int` `>& a,` ` ` `vector<` `int` `>& jump, ` `int` `n)` `{` ` ` `// Initialise Inverse Graph` ` ` `vector<` `int` `> invGr[n + 1];` ` ` `// Stores the result for each index` ` ` `vector<` `int` `> ans(n + 1, -1);` ` ` `for` `(` `int` `i = 1; i <= n; i++) {` ` ` `// For the jumped index` ` ` `for` `(` `int` `ind : { i + jump[i],` ` ` `i - jump[i] }) {` ` ` `// If the ind is valid then` ` ` `// add reverse directed edge` ` ` `if` `(ind >= 1 and ind <= n) {` ` ` `invGr[ind].push_back(i);` ` ` `}` ` ` `}` ` ` `}` ` ` `// Multi-source bfs with odd` ` ` `// numbers as source by passing 0` ` ` `bfs(n, a, invGr, ans, 0);` ` ` `// Multi-source bfs with even` ` ` `// numbers as source by passing 1` ` ` `bfs(n, a, invGr, ans, 1);` ` ` `// Print the answer` ` ` `for` `(` `int` `i = 1; i <= n; i++) {` ` ` `cout << ans[i] << ` `' '` `;` ` ` `}` `}` `// Driver Code` `int` `main()` `{` ` ` `vector<` `int` `> arr = { 0, 4, 2, 5, 2, 1 };` ` ` `vector<` `int` `> jump = { 0, 1, 2, 3, 1, 2 };` ` ` `int` `N = arr.size();` ` ` `minJumps(arr, jump, N - 1);` ` ` `return` `0;` `}` |

## Java

`// Java program for above approach` `import` `java.util.*;` `import` `java.lang.*;` `class` `Gfg` `{` ` ` `// Bfs for odd numbers are source` ` ` `static` `void` `bfs(` `int` `n, ` `int` `[] a,` ` ` `ArrayList<ArrayList<Integer>> invGr,` ` ` `int` `[] ans, ` `int` `parity)` ` ` `{` ` ` `// Initialize queue` ` ` `Queue<Integer> q = ` `new` `LinkedList<>();` ` ` `// Stores for each node, the nodes` ` ` `// visited and their distances` ` ` `int` `[] vis = ` `new` `int` `[n + ` `1` `];` ` ` `int` `[] dist = ` `new` `int` `[n + ` `1` `];` ` ` `// Push odd and even numbers` ` ` `// as sources to the queue` ` ` `// If parity is 0 -> odd` ` ` `// Otherwise -> even` ` ` `for` `(` `int` `i = ` `1` `; i <= n; i++) {` ` ` `if` `(((a[i] + parity) & ` `1` `) != ` `0` `) {` ` ` `q.add(i);` ` ` `vis[i] = ` `1` `;` ` ` `}` ` ` `}` ` ` `// Perform multi-source bfs` ` ` `while` `(!q.isEmpty()) {` ` ` `// Extract the front element` ` ` `// of the queue` ` ` `int` `v = q.peek();` ` ` `q.poll();` ` ` `// Traverse nodes connected` ` ` `// to the current node` ` ` `for` `(Integer u : invGr.get(v)) {` ` ` `// If u is not visited` ` ` `if` `(vis[u] == ` `0` `) {` ` ` `dist[u] = dist[v] + ` `1` `;` ` ` `vis[u] = ` `1` `;` ` ` `// If element with opposite` ` ` `// parity is obtained` ` ` `if` `((a[u] + parity) % ` `2` `== ` `0` `) {` ` ` `if` `(ans[u] == -` `1` `)` ` ` `// Store its distance from` ` ` `// source in ans[]` ` ` `ans[u] = dist[u];` ` ` `}` ` ` `// Push the current neighbour` ` ` `// to the queue` ` ` `q.add(u);` ` ` `}` ` ` `}` ` ` `}` ` ` `}` ` ` `// Function to find the minimum jumps` ` ` `// required by each index to reach` ` ` `// element of opposite parity` ` ` `static` `void` `minJumps(` `int` `[] a,` ` ` `int` `[] jump, ` `int` `n)` ` ` `{` ` ` `// Initialise Inverse Graph` ` ` `ArrayList<ArrayList<Integer>> invGr = ` `new` `ArrayList<>();` ` ` `for` `(` `int` `i = ` `0` `; i <= n; i++)` ` ` `invGr.add(` `new` `ArrayList<Integer>());` ` ` `// Stores the result for each index` ` ` `int` `[] ans = ` `new` `int` `[n + ` `1` `];` ` ` `Arrays.fill(ans, -` `1` `);` ` ` `for` `(` `int` `i = ` `1` `; i <= n; i++)` ` ` `{` ` ` `// For the jumped index` ` ` `// If the ind is valid then` ` ` `// add reverse directed edge` ` ` `if` `(i+ jump[i] >= ` `1` `&& i+jump[i] <= n) {` ` ` `invGr.get(i+ jump[i]).add(i);` ` ` `}` ` ` `if` `(i-jump[i] >= ` `1` `&& i-jump[i] <= n) {` ` ` `invGr.get(i- jump[i]).add(i);` ` ` `}` ` ` `}` ` ` `// Multi-source bfs with odd` ` ` `// numbers as source by passing 0` ` ` `bfs(n, a, invGr, ans, ` `0` `);` ` ` `// Multi-source bfs with even` ` ` `// numbers as source by passing 1` ` ` `bfs(n, a, invGr, ans, ` `1` `);` ` ` `// Print the answer` ` ` `for` `(` `int` `i = ` `1` `; i <= n; i++)` ` ` `{` ` ` `System.out.print(ans[i] + ` `" "` `);` ` ` `}` ` ` `}` ` ` `// Driver function` ` ` `public` `static` `void` `main (String[] args)` ` ` `{` ` ` `int` `[] arr = { ` `0` `, ` `4` `, ` `2` `, ` `5` `, ` `2` `, ` `1` `};` ` ` `int` `[] jump = { ` `0` `, ` `1` `, ` `2` `, ` `3` `, ` `1` `, ` `2` `};` ` ` `int` `N = arr.length;` ` ` `minJumps(arr, jump, N - ` `1` `);` ` ` `}` `}` `// This code is contributed by offbeat` |

## Python3

`# Python3 program for the above approach` ` ` `# Bfs for odd numbers are source` `def` `bfs(n, a, invGr, ans, parity):` ` ` ` ` `# Initialize queue` ` ` `q ` `=` `[]` ` ` ` ` `# Stores for each node, the nodes` ` ` `# visited and their distances` ` ` `vis ` `=` `[` `0` `for` `i ` `in` `range` `(n ` `+` `1` `)]` ` ` `dist ` `=` `[` `0` `for` `i ` `in` `range` `(n ` `+` `1` `)]` ` ` ` ` `# Push odd and even numbers` ` ` `# as sources to the queue` ` ` ` ` `# If parity is 0 -> odd` ` ` `# Otherwise -> even` ` ` `for` `i ` `in` `range` `(` `1` `, n ` `+` `1` `):` ` ` `if` `((a[i] ` `+` `parity) & ` `1` `):` ` ` `q.append(i)` ` ` `vis[i] ` `=` `1` ` ` ` ` `# Perform multi-source bfs` ` ` `while` `(` `len` `(q) !` `=` `0` `):` ` ` ` ` `# Extract the front element` ` ` `# of the queue` ` ` `v ` `=` `q[` `0` `]` ` ` `q.pop(` `0` `)` ` ` ` ` `# Traverse nodes connected` ` ` `# to the current node` ` ` `for` `u ` `in` `invGr[v]:` ` ` ` ` `# If u is not visited` ` ` `if` `(` `not` `vis[u]):` ` ` `dist[u] ` `=` `dist[v] ` `+` `1` ` ` `vis[u] ` `=` `1` ` ` ` ` `# If element with opposite` ` ` `# parity is obtained` ` ` `if` `((a[u] ` `+` `parity) ` `%` `2` `=` `=` `0` `):` ` ` `if` `(ans[u] ` `=` `=` `-` `1` `):` ` ` ` ` `# Store its distance from` ` ` `# source in ans[]` ` ` `ans[u] ` `=` `dist[u]` ` ` ` ` `# Push the current neighbour` ` ` `# to the queue` ` ` `q.append(u)` ` ` `# Function to find the minimum jumps` `# required by each index to reach` `# element of opposite parity` `def` `minJumps(a, jump, n):` ` ` `# Initialise Inverse Graph` ` ` `invGr ` `=` `[[] ` `for` `i ` `in` `range` `(n ` `+` `1` `)]` ` ` ` ` `# Stores the result for each index` ` ` `ans ` `=` `[` `-` `1` `for` `i ` `in` `range` `(n ` `+` `1` `)]` ` ` ` ` `for` `i ` `in` `range` `(` `1` `, n ` `+` `1` `):` ` ` ` ` `# For the jumped index` ` ` `for` `ind ` `in` `[i ` `+` `jump[i], i ` `-` `jump[i]]:` ` ` ` ` `# If the ind is valid then` ` ` `# add reverse directed edge` ` ` `if` `(ind >` `=` `1` `and` `ind <` `=` `n):` ` ` `invGr[ind].append(i)` ` ` ` ` `# Multi-source bfs with odd` ` ` `# numbers as source by passing 0` ` ` `bfs(n, a, invGr, ans, ` `0` `)` ` ` ` ` `# Multi-source bfs with even` ` ` `# numbers as source by passing 1` ` ` `bfs(n, a, invGr, ans, ` `1` `)` ` ` ` ` `# Print the answer` ` ` `for` `i ` `in` `range` `(` `1` `, n ` `+` `1` `):` ` ` `print` `(` `str` `(ans[i]), end ` `=` `' '` `)` ` ` `# Driver Code` `if` `__name__` `=` `=` `'__main__'` `:` ` ` ` ` `arr ` `=` `[ ` `0` `, ` `4` `, ` `2` `, ` `5` `, ` `2` `, ` `1` `]` ` ` `jump ` `=` `[ ` `0` `, ` `1` `, ` `2` `, ` `3` `, ` `1` `, ` `2` `]` ` ` ` ` `N ` `=` `len` `(arr)` ` ` ` ` `minJumps(arr, jump, N ` `-` `1` `)` ` ` `# This code is contributed by pratham76` |

## Javascript

`<script>` `// JavaScript program for the above approach` `// Bfs for odd numbers are source` `function` `bfs(n, a, invGr, ans, parity)` `{` ` ` `// Initialize queue` ` ` `var` `q = [];` ` ` `// Stores for each node, the nodes` ` ` `// visited and their distances` ` ` `var` `vis = Array(n+1).fill(0);` ` ` `var` `dist = Array(n+1).fill(0);` ` ` `// Push odd and even numbers` ` ` `// as sources to the queue` ` ` `// If parity is 0 -> odd` ` ` `// Otherwise -> even` ` ` `for` `(` `var` `i = 1; i <= n; i++) {` ` ` `if` `((a[i] + parity) & 1) {` ` ` `q.push(i);` ` ` `vis[i] = 1;` ` ` `}` ` ` `}` ` ` `// Perform multi-source bfs` ` ` `while` `(q.length!=0) {` ` ` `// Extract the front element` ` ` `// of the queue` ` ` `var` `v = q[0];` ` ` `q.shift();` ` ` `// Traverse nodes connected` ` ` `// to the current node` ` ` `invGr[v].forEach(u => {` ` ` ` ` `// If u is not visited` ` ` `if` `(!vis[u]) {` ` ` `dist[u] = dist[v] + 1;` ` ` `vis[u] = 1;` ` ` `// If element with opposite` ` ` `// parity is obtained` ` ` `if` `((a[u] + parity) % 2 == 0) {` ` ` `if` `(ans[u] == -1)` ` ` `// Store its distance from` ` ` `// source in ans[]` ` ` `ans[u] = dist[u];` ` ` `}` ` ` `// Push the current neighbour` ` ` `// to the queue` ` ` `q.push(u);` ` ` `}` ` ` `});` ` ` `}` ` ` `return` `ans;` `}` `// Function to find the minimum jumps` `// required by each index to reach` `// element of opposite parity` `function` `minJumps(a, jump, n)` `{` ` ` `// Initialise Inverse Graph` ` ` `var` `invGr = Array.from(Array(n+1), ()=>Array())` ` ` `// Stores the result for each index` ` ` `var` `ans = Array(n + 1).fill(-1);` ` ` `for` `(` `var` `i = 1; i <= n; i++) {` ` ` `// For the jumped index` ` ` `[i + jump[i], i - jump[i]].forEach(ind => {` ` ` ` ` `// If the ind is valid then` ` ` `// add reverse directed edge` ` ` `if` `(ind >= 1 && ind <= n) {` ` ` `invGr[ind].push(i);` ` ` `}` ` ` `});` ` ` `}` ` ` `// Multi-source bfs with odd` ` ` `// numbers as source by passing 0` ` ` `ans = bfs(n, a, invGr, ans, 0);` ` ` `// Multi-source bfs with even` ` ` `// numbers as source by passing 1` ` ` `ans = bfs(n, a, invGr, ans, 1);` ` ` `// Print the answer` ` ` `for` `(` `var` `i = 1; i <= n; i++) {` ` ` `document.write( ans[i] + ` `' '` `);` ` ` `}` `}` `// Driver Code` `var` `arr = [0, 4, 2, 5, 2, 1];` `var` `jump = [0, 1, 2, 3, 1, 2];` `var` `N = arr.length;` `minJumps(arr, jump, N - 1);` `</script>` |

**Output:**

3 2 -1 1 -1

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

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