Given two arrays arr1[] and arr2[], both of size N and an integer X, the task is to check if the sum of same-indexed elements of both the arrays at corresponding indices can be made at most K after rearranging the second array. If it is possible then print “Yes” else print “No”.
Examples:
Input: arr1[] = {1, 2, 3}, arr2[] = {1, 1, 2}, X = 4
Output: Yes
Explanation:
Rearranging the array B[] as {1, 2, 1}. Now the sum of corresponding indices are:
A[0] + B[0] = 1 + 1 = 2 ? 4
A[1] + B[1] = 2 + 2 = 4 ? 4
A[2] + B[2] = 3 + 1 = 4 ? 4Input: arr1[] = {1, 2, 3, 4}, arr2[] = {1, 2, 3, 4}, X = 4
Output: No
Explanation: There is no way that the array B[] can be rearranged such that the condition A[i] + B[i] <= X is satisfied.
Naive Approach: The simplest approach is to generate all possible permutations of the array B[] and if any permutation satisfies the given condition, then print Yes. Otherwise, print No.
Time Complexity: O(N!)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to use Sorting. Follow the steps below to solve the problem:
- Sort the array arr1[] in ascending order and arr2[] in descending order.
- Now, traverse both the arrays simultaneously and if there exists any pair such that the sum of arr1[i] and arr2[] exceeds X, then print “No”. Otherwise, print “Yes”.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to check if elements// of B[] can be rearranged// such that A[i] + B[i] <= Xvoid rearrange(int A[], int B[], int N, int X){ // Checks the given condition bool flag = true; // Sort A[] in ascending order sort(A, A + N); // Sort B[] in descending order sort(B, B + N, greater<int>()); // Traverse the arrays A[] and B[] for (int i = 0; i < N; i++) { // If A[i] + B[i] exceeds X if (A[i] + B[i] > X) { // Rearrangement not possible, // set flag to false flag = false; break; } } // If flag is true if (flag) cout << "Yes"; // Otherwise else cout << "No";}// Driver Codeint main(){ int A[] = { 1, 2, 3 }; int B[] = { 1, 1, 2 }; int X = 4; int N = sizeof(A) / sizeof(A[0]); // Function Call rearrange(A, B, N, X); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG { // Function to check if elements // of B[] can be rearranged // such that A[i] + B[i] <= X static void rearrange(int A[], int B[], int N, int X) { // Checks the given condition boolean flag = true; // Sort A[] in ascending order Arrays.sort(A); // Sort B[] in descending order Arrays.sort(B); // Traverse the arrays A[] and B[] for (int i = 0; i < N; i++) { // If A[i] + B[i] exceeds X if (A[i] + B[N - 1 - i] > X) { // Rearrangement not possible, // set flag to false flag = false; break; } } // If flag is true if (flag == true) System.out.print("Yes"); // Otherwise else System.out.print("No"); } // Driver Code public static void main (String[] args) { int A[] = { 1, 2, 3 }; int B[] = { 1, 1, 2 }; int X = 4; int N = A.length; // Function Call rearrange(A, B, N, X); }}// This code is contributed by AnkThon |
Python3
# Python3 program for the above approach# Function to check if elements# of B can be rearranged# such that A[i] + B[i] <= Xdef rearrange(A, B, N, X): # Checks the given condition flag = True # Sort A in ascending order A = sorted(A) # Sort B in descending order B = sorted(B)[::-1] # Traverse the arrays A and B for i in range(N): # If A[i] + B[i] exceeds X if (A[i] + B[i] > X): # Rearrangement not possible, # set flag to false flag = False break # If flag is true if (flag): print("Yes") # Otherwise else: print("No")# Driver Codeif __name__ == '__main__': A = [ 1, 2, 3 ] B = [ 1, 1, 2 ] X = 4 N = len(A) # Function Call rearrange(A, B, N, X)# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;using System.Collections.Generic; class GFG { // Function to check if elements // of B[] can be rearranged // such that A[i] + B[i] <= X static void rearrange(int[] A, int[] B, int N, int X) { // Checks the given condition bool flag = true; // Sort A[] in ascending order Array.Sort(A); // Sort B[] in descending order Array.Sort(B); // Traverse the arrays A[] and B[] for (int i = 0; i < N; i++) { // If A[i] + B[i] exceeds X if (A[i] + B[N - 1 - i] > X) { // Rearrangement not possible, // set flag to false flag = false; break; } } // If flag is true if (flag == true) Console.WriteLine("Yes"); // Otherwise else Console.WriteLine("No"); } // Driver Code public static void Main() { int []A = { 1, 2, 3 }; int []B = { 1, 1, 2 }; int X = 4; int N = A.Length; // Function Call rearrange(A, B, N, X); }}// This code is contributed by AnkThon |
Javascript
<script>// Javascript program of the above approach// Function to check if elements// of B[] can be rearranged// such that A[i] + B[i] <= Xfunction rearrange(A, B, N, X){ // Checks the given condition let flag = true; // Sort A[] in ascending order A.sort(); // Sort B[] in descending order B.sort(); // Traverse the arrays A[] and B[] for(let i = 0; i < N; i++) { // If A[i] + B[i] exceeds X if (A[i] + B[N - 1 - i] > X) { // Rearrangement not possible, // set flag to false flag = false; break; } } // If flag is true if (flag == true) document.write("Yes"); // Otherwise else document.write("No");}// Driver Codelet A = [ 1, 2, 3 ];let B = [ 1, 1, 2 ];let X = 4;let N = A.length;// Function Callrearrange(A, B, N, X);// This code is contributed by avijitmondal1998</script> |
Output:
Yes
Time Complexity: O(N*log N)
Auxiliary Space: O(1)
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