Permutation of an array that has smaller values from another array
Last Updated :
18 Nov, 2022
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 permutation
void 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 program
int 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 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 permutation
static 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 Code
public 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 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 Code
if __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
// Include namespace system
using 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.
<script>
// JavaScript program to find permutation of
// an array that has smaller values from
// another array
// Function to document.write required permutation
function 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.length
anyPermutation(A, B, n)
// This code is contributed by shinjanpatra
</script>
Time Complexity: O(N*log(N)), where N is the length of array.
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