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

• Difficulty Level : Easy
• Last Updated : 16 Nov, 2022

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 to print “Yes” if any such permutation exists, otherwise print “No”.

Examples :

```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```
Recommended Practice

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 that 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 the implementation is this approach:

## C++

 `// C++ program to check whether permutation of two` `// arrays satisfy the condition a[i] + b[i] >= k.` `#include ` `using` `namespace` `std;`   `// Check whether 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;` `}`   `// This code is contributed by Aditya Kumar (adityakumar129)`

## C

 `// C program to check whether permutation of two` `// arrays satisfy the condition a[i] + b[i] >= k.` `#include ` `#include ` `#include `   `// Compare function for qsort for Increasing Order` `int` `cmpfunc1(``const` `void``* a, ``const` `void``* b)` `{` `    ``return` `(*(``int``*)a - *(``int``*)b);` `}`   `// Compare function for qsort for decreasing Order` `int` `cmpfunc2(``const` `void``* a, ``const` `void``* b)` `{` `    ``return` `(*(``int``*)b - *(``int``*)a);` `}`   `// Check whether any permutation exists which` `// satisfy the condition.` `bool` `isPossible(``int` `a[], ``int` `b[], ``int` `n, ``int` `k)` `{` `    ``// Sort the array a[] in decreasing order.` `    ``qsort``(a, n, ``sizeof``(``int``), cmpfunc1);`   `    ``// Sort the array b[] in increasing order.` `    ``qsort``(b, n, ``sizeof``(``int``), cmpfunc2);`   `    ``// 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) ? ``printf``(``"Yes"``) : ``printf``(``"No"``);` `    ``return` `0;` `}`   `// This code is contributed by Aditya Kumar (adityakumar129)`

## Java

 `// Java program to check whether ` `// permutation of two arrays satisfy` `// the condition a[i] + b[i] >= k.` `import` `java.util.*;`   `class` `GFG ` `{` `// Check whether any permutation ` `// exists which satisfy the condition.` `static` `boolean` `isPossible(Integer a[], ``int` `b[],` `                                  ``int` `n, ``int` `k) ` `{` `    ``// Sort the array a[] in decreasing order.` `    ``Arrays.sort(a, Collections.reverseOrder());`   `    ``// Sort the array b[] in increasing order.` `    ``Arrays.sort(b);`   `    ``// Checking condition on each index.` `    ``for` `(``int` `i = ``0``; i < n; i++)` `    ``if` `(a[i] + b[i] < k)` `        ``return` `false``;`   `    ``return` `true``;` `}`   `// Driver code` `public` `static` `void` `main(String[] args) {` `    ``Integer a[] = {``2``, ``1``, ``3``};` `    ``int` `b[] = {``7``, ``8``, ``9``};` `    ``int` `k = ``10``;` `    ``int` `n = a.length;`   `    ``if` `(isPossible(a, b, n, k))` `    ``System.out.print(``"Yes"``);` `    ``else` `    ``System.out.print(``"No"``);` `}` `}`   `// This code is contributed by Anant Agarwal.`

## Python3

 `# Python program to check` `# whether permutation of two` `# arrays satisfy the condition` `# a[i] + b[i] >= k.`   `# Check whether any` `# permutation exists which` `# satisfy the condition.` `def` `isPossible(a,b,n,k):`   `    ``# Sort the array a[]` `    ``# in decreasing order.` `    ``a.sort(reverse``=``True``)` ` `  `    ``# Sort the array b[]` `    ``# in increasing order.` `    ``b.sort()` ` `  `    ``# Checking condition` `    ``# on each index.` `    ``for` `i ``in` `range``(n):` `        ``if` `(a[i] ``+` `b[i] < k):` `            ``return` `False` ` `  `    ``return` `True`     `# Driver code`   `a ``=` `[ ``2``, ``1``, ``3``]` `b ``=` `[``7``, ``8``, ``9``]` `k ``=` `10` `n ``=``len``(a)` ` `  `if``(isPossible(a, b, n, k)):` `    ``print``(``"Yes"``)` `else``:` `    ``print``(``"No"``)`   `# This code is contributed` `# by Anant Agarwal.`

## C#

 `// C# program to check whether ` `// permutation of two arrays satisfy` `// the condition a[i] + b[i] >= k.` `using` `System;`   `class` `GFG ` `{` `// Check whether any permutation ` `// exists which satisfy the condition.` `static` `bool` `isPossible(``int` `[]a, ``int` `[]b,` `                       ``int` `n, ``int` `k) ` `{` `    ``// Sort the array a[] ` `    ``// in decreasing order.` `    ``Array.Sort(a);`   `    ``// Sort the array b[] ` `    ``// in increasing order.` `    ``Array.Reverse(b);`   `    ``// Checking condition on each index.` `    ``for` `(``int` `i = 0; i < n; i++)` `    ``if` `(a[i] + b[i] < k)` `        ``return` `false``;`   `    ``return` `true``;` `}`   `// Driver code` `public` `static` `void` `Main() ` `{` `    ``int` `[]a = {2, 1, 3};` `    ``int` `[]b = {7, 8, 9};` `    ``int` `k = 10;` `    ``int` `n = a.Length;`   `    ``if` `(isPossible(a, b, n, k))` `    ``Console.WriteLine(``"Yes"``);` `    ``else` `    ``Console.WriteLine(``"No"``);` `}` `}`   `// This code is contributed by anuj_67.`

## PHP

 `= k.`   `// Check whether any permutation ` `// exists which satisfy the condition.` `function` `isPossible( ``\$a``, ``\$b``, ``\$n``, ``\$k``)` `{` `    `  `    ``// Sort the array a[] in` `    ``// decreasing order.` `    ``sort(``\$a``);`   `    ``// Sort the array b[] in ` `    ``// increasing order.` `    ``rsort(``\$b``);`   `    ``// Checking condition on each` `    ``// index.` `    ``for` `( ``\$i` `= 0; ``\$i` `< ``\$n``; ``\$i``++)` `        ``if` `(``\$a``[``\$i``] + ``\$b``[``\$i``] < ``\$k``)` `            ``return` `false;`   `    ``return` `true;` `}`   `// Driven Program` `    ``\$a` `= ``array``( 2, 1, 3 );` `    ``\$b` `= ``array``( 7, 8, 9 );` `    ``\$k` `= 10;` `    ``\$n` `= ``count``(``\$a``);`   `    ``if``(isPossible(``\$a``, ``\$b``, ``\$n``, ``\$k``)) ` `        ``echo` `"Yes"` `;` `    ``else` `        ``echo` `"No"``;`   `// This code is contributed by ` `// anuj_67.` `?>`

## Javascript

 ``

Output

`Yes`

Time Complexity: O(n log n).
Auxiliary Space: O(1)
This approach is contributed by Anuj Chauhan

Method 2:
In the above case, we were just sorting the elements and checking the statement for n time (making pairs of ith index) resulting in time complexity of O(n log n).

As we can see we only need to check that is there any pair whose sum is greater than K. To solve it we have to just find the maximum pair sum.

Follow the steps below to implement the above idea:

1. We will first find the max element from the first array and then the max element from the second array.
2. Check if the sum of both max elements is greater than K
• If true, return true
• otherwise false.

Below is the implementation of the above approach:

## C++

 `#include` `using` `namespace` `std;`   `bool` `isPossible(``long` `long` `arr1[] , ``long` `long` `arr2[] , ``int` `n , ``long` `long` `K){` `    ``//FINDING THE MAX ELEMENT`   `    ``long` `long` `a = *max_element(arr1 , arr1+n); ``// *max_element is STL function used to find the max element from the array.` `    ``long` `long` `b = *max_element(arr2 , arr2+n);`   `    ``long` `long` `ans = a+b;`   `    ``// CHECKING THE CONDITION` `    ``return` `ans >= K;` `}`   `int` `main(){` `    ``long` `long` `arr1[] = {2,1,3};` `    ``long` `long` `arr2[] = {7,8,9};`   `    ``long` `long` `k = 5;`   `    ``int` `n = ``sizeof``(arr1)/``sizeof``(arr1[0]);`   `    ``if` `(isPossible(arr1 , arr2 , n , k))` `    ``{` `        ``cout<<``"Yes"``<

## Java

 `// Java code for the above approach`   `import` `java.io.*;` `import` `java.util.*;`   `class` `GFG {`   `  ``static` `boolean` `isPossible(``int``[] arr1, ``int``[] arr2, ``int` `n,` `                            ``int` `K)` `  ``{` `    ``int` `a = Arrays.stream(arr1).max().getAsInt();` `    ``int` `b = Arrays.stream(arr2).max().getAsInt();`   `    ``int` `ans = a + b;`   `    ``// CHECKING THE CONDITION` `    ``return` `ans >= K;` `  ``}`   `  ``public` `static` `void` `main(String[] args)` `  ``{` `    ``int``[] arr1 = { ``2``, ``1``, ``3` `};` `    ``int``[] arr2 = { ``7``, ``8``, ``9` `};`   `    ``int` `k = ``5``;`   `    ``int` `n = arr1.length;`   `    ``if` `(isPossible(arr1, arr2, n, k)) {` `      ``System.out.println(``"Yes"``);` `    ``}` `    ``else` `{` `      ``System.out.println(``"No"``);` `    ``}` `  ``}` `}`   `// This code is contributed by lokesh`

Output

`Yes`

Time Complexity: O(n)
Auxiliary Space: O(1)

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