Given two arrays A and B of equal size. The task is to print any permutation of array A such that the number of indices i for which A[i] > B[i] is maximized.
Examples:
Input: A = [12, 24, 8, 32],
B = [13, 25, 32, 11]
Output: 24 32 8 12
Input: A = [2, 7, 11, 15],
B = [1, 10, 4, 11]
Output: 2 11 7 15
If the smallest element in A beats the smallest element in B, we should pair them. Otherwise, it is useless for our score, as it can’t beat any other element of B.
With above strategy we make two vector of pairs, Ap for A and Bp for B with their element and respective index. Then sort both vectors and simulate them. Whenever we found any element in vector Ap such that Ap[i].first > Bp[j].first for some (i, j) we pair them i:e we update our answer array to ans[Bp[j].second] = Ap[i].first. However if Ap[i].first < Bp[j].first for some (i, j) then we store them in vector remain and finally pair them with any one.
Below is the implementation of above approach:
C++
// C++ program to find permutation of an array that// has smaller values from another array#include <bits/stdc++.h>using namespace std;// Function to print required permutationvoid anyPermutation(int A[], int B[], int n){ // Storing elements and indexes vector<pair<int, int> > Ap, Bp; for (int i = 0; i < n; i++) Ap.push_back(make_pair(A[i], i)); for (int i = 0; i < n; i++) Bp.push_back(make_pair(B[i], i)); sort(Ap.begin(), Ap.end()); sort(Bp.begin(), Bp.end()); int i = 0, j = 0, ans[n] = { 0 }; // Filling the answer array vector<int> remain; while (i < n && j < n) { // pair element of A and B if (Ap[i].first > Bp[j].first) { ans[Bp[j].second] = Ap[i].first; i++; j++; } else { remain.push_back(i); i++; } } // Fill the remaining elements of answer j = 0; for (int i = 0; i < n; ++i) if (ans[i] == 0) { ans[i] = Ap[remain[j]].first; j++; } // Output required permutation for (int i = 0; i < n; ++i) cout << ans[i] << " ";}// Driver programint main(){ int A[] = { 12, 24, 8, 32 }; int B[] = { 13, 25, 32, 11 }; int n = sizeof(A) / sizeof(A[0]); anyPermutation(A, B, n); return 0;}// This code is written by Sanjit_Prasad |
Java
// Java program to find permutation of an // array that has smaller values from // another array import java.io.*;import java.lang.*;import java.util.*;class GFG{// Function to print required permutationstatic void anyPermutation(int A[], int B[], int n){ // Storing elements and indexes ArrayList<int[]> Ap = new ArrayList<>(); ArrayList<int[]> Bp = new ArrayList<>(); for(int i = 0; i < n; i++) Ap.add(new int[] { A[i], i }); for(int i = 0; i < n; i++) Bp.add(new int[] { B[i], i }); // Sorting the Both Ap and Bp Collections.sort(Ap, (x, y) -> { if (x[0] != y[0]) return x[0] - y[0]; return y[1] - y[1]; }); Collections.sort(Bp, (x, y) -> { if (x[0] != y[0]) return x[0] - y[0]; return y[1] - y[1]; }); int i = 0, j = 0; int ans[] = new int[n]; // Filling the answer array ArrayList<Integer> remain = new ArrayList<>(); while (i < n && j < n) { // Pair element of A and B if (Ap.get(i)[0] > Bp.get(j)[0]) { ans[Bp.get(j)[1]] = Ap.get(i)[0]; i++; j++; } else { remain.add(i); i++; } } // Fill the remaining elements of answer j = 0; for(i = 0; i < n; ++i) if (ans[i] == 0) { ans[i] = Ap.get(remain.get(j))[0]; j++; } // Output required permutation for(i = 0; i < n; ++i) System.out.print(ans[i] + " ");}// Driver Codepublic static void main(String[] args){ int A[] = { 12, 24, 8, 32 }; int B[] = { 13, 25, 32, 11 }; int n = A.length; anyPermutation(A, B, n);}}// This code is contributed by Kingash |
Python3
# Python3 program to find permutation of # an array that has smaller values from# another array # Function to print required permutation def anyPermutation(A, B, n): # Storing elements and indexes Ap, Bp = [], [] for i in range(0, n): Ap.append([A[i], i]) for i in range(0, n): Bp.append([B[i], i]) Ap.sort() Bp.sort() i, j = 0, 0, ans = [0] * n # Filling the answer array remain = [] while i < n and j < n: # pair element of A and B if Ap[i][0] > Bp[j][0]: ans[Bp[j][1]] = Ap[i][0] i += 1 j += 1 else: remain.append(i) i += 1 # Fill the remaining elements # of answer j = 0 for i in range(0, n): if ans[i] == 0: ans[i] = Ap[remain[j]][0] j += 1 # Output required permutation for i in range(0, n): print(ans[i], end = " ") # Driver Codeif __name__ == "__main__": A = [ 12, 24, 8, 32 ] B = [ 13, 25, 32, 11 ] n = len(A) anyPermutation(A, B, n) # This code is contributed # by Rituraj Jain |
C#
// Include namespace systemusing System;using System.Collections.Generic;public class GFG{ // Function to print required permutation public static void anyPermutation(int[] A, int[] B, int n) { // Storing elements and indexes List<int[]> Ap = new List<int[]>(); List<int[]> Bp = new List<int[]>(); for (int i = 0; i < n; i++) { Ap.Add(new int[]{A[i], i}); } for (int i = 0; i < n; i++) { Bp.Add(new int[]{B[i], i}); } // Sorting the Both Ap and Bp Ap.Sort((x,y)=> (x[0] != y[0]) ? x[0] - y[0] : y[1] - y[1]); Bp.Sort((x,y)=> (x[0] != y[0]) ? x[0] - y[0] : y[1] - y[1]); var ii = 0; var j = 0; int[] ans = new int[n]; // Filling the answer array var remain = new List<int>(); while (ii < n && j < n) { // Pair element of A and B if (Ap[ii][0] > Bp[j][0]) { ans[Bp[j][1]] = Ap[ii][0]; ii++; j++; } else { remain.Add(ii); ii++; } } // Fill the remaining elements of answer j = 0; for (var i = 0; i < n; ++i) { if (ans[i] == 0) { ans[i] = Ap[remain[j]][0]; j++; } } // Output required permutation for (var i = 0; i < n; ++i) { Console.Write(ans[i].ToString() + " "); } } // Driver Code public static void Main(String[] args) { int[] A = {12, 24, 8, 32}; int[] B = {13, 25, 32, 11}; var n = A.Length; GFG.anyPermutation(A, B, n); }}// This code is contributed by aadityaburujwale. |
Javascript
<script>// JavaScript program to find permutation of// an array that has smaller values from// another array// Function to document.write required permutationfunction anyPermutation(A, B, n){ // Storing elements and indexes let Ap = [], Bp = [] for(let i=0;i<n;i++){ Ap.push([A[i], i]) } for(let i=0;i<n;i++){ Bp.push([B[i], i]) } Ap.sort() Bp.sort() let i = 0 let j = 0 let ans = new Array(n).fill(0) // Filling the answer array let remain = [] while(i < n && j < n){ // pair element of A and B if(Ap[i][0] > Bp[j][0]){ ans[Bp[j][1]] = Ap[i][0] i += 1 j += 1 } else{ remain.push(i) i += 1 } } // Fill the remaining elements // of answer j = 0 for(let i=0;i<n;i++){ if(ans[i] == 0){ ans[i] = Ap[remain[j]][0] j += 1 } } // Output required permutation for(let i=0;i<n;i++){ document.write(ans[i]," ") }}// Driver Code let A = [ 12, 24, 8, 32 ]let B = [ 13, 25, 32, 11 ]let n = A.lengthanyPermutation(A, B, n) // This code is contributed by shinjanpatra</script> |
Output
24 32 8 12
Time Complexity: O(N*log(N)), where N is the length of array.
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