Permutation of an array that has smaller values from another array

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++

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// CPP 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

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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 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

Output:

24 32 8 12

Time Complexity: O(N*log(N)), where N is the length of array.



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Improved By : rituraj_jain