Permute two arrays such that sum of every pair is greater or equal to K


Given two arrays of equal size n and an integer k. The task is to permute both arrays such that sum of their corresponding element is greater than or equal to k i.e a[i] + b[i] >= k. The task is print “Yes” if any such permutation exists, otherwise print “No”.


Input : a[] = {2, 1, 3}, 
        b[] = { 7, 8, 9 }, 
        k = 10. 
Output : Yes
Permutation  a[] = { 1, 2, 3 } and b[] = { 9, 8, 7 } 
satisfied the condition a[i] + b[i] >= K.

Input : a[] = {1, 2, 2, 1}, 
        b[] = { 3, 3, 3, 4 }, 
        k = 5. 
Output : No

The idea is to sort one array in ascending order and another array in descending order and if any index does not satisfy the condition a[i] + b[i] >= K then print “No”, else print “Yes”.

If the condition fails on sorted arrays, then there exists no permutation of arrays which can satisfy the inequality. Proof,

Assume asort[] be sorted a[] in ascending order and bsort[] be sorted b[] in descending order.
Let new permutation b[] is created by swapping any two indices i, j of bsort[],

  • Case 1: i < j and element at b[i] is now bsort[j].
    Now, asort[i] + bsort[j] < K, because bsort[i] > bsort[j] as b[] is sorted in decreasing order and we know asort[i] + bsort[i] < k.
  • Case 2: i > j and element at b[i] is now bsort[j].
    Now, asort[j] + bsort[i] < k, because asort[i] > asort[j] as a[] is sorted in increasing order and we know asort[i] + bsort[i] < k.

Below is C++ implementation is this approach:

// C++ program to check whether permutation of two
// arrays satisfy the condition a[i] + b[i] >= k.
using namespace std;

// Check wheather any permutation exists which
// satisfy the condition.
bool isPossible(int a[], int b[], int n, int k)
    // Sort the array a[] in decreasing order.
    sort(a, a + n);

    // Sort the array b[] in increasing order.
    sort(b, b + n, greater<int>());

    // Checking condition on each index.
    for (int i = 0; i < n; i++)
        if (a[i] + b[i] < k)
            return false;

    return true;

// Driven Program
int main()
    int a[] = { 2, 1, 3 };
    int b[] = { 7, 8, 9 };
    int k = 10;
    int n = sizeof(a)/sizeof(a[0]);

    isPossible(a, b, n, k) ? cout << "Yes" :
                             cout << "No";
    return 0;



Time Complexity: O(n log n).

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

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