# Split array into minimum number of subsets having maximum pair sum at most K

• Difficulty Level : Medium
• Last Updated : 19 Jul, 2021

Given an array, arr[] of size N and an integer K, the task is to partition the array into the minimum number of subsets such that the maximum pair sum in each subset is less than or equal to K.

Examples:

Input: arr[] = {1, 2, 3, 4, 5}, K = 5
Output: 3
Explanation:
Subset having maximum pair sum less than or equal to K(= 5) are: {{2, 3}, {1, 4}, {5}}.
Therefore, the required output is 3.

Input: arr[] = {2, 6, 8, 10, 20, 25}, K = 26
Output: 3
Explanation:
Subset having maximum pair sum less than or equal to K(=26) are: {{2, 6, 8, 10}, {20}, {25}}.
Therefore, the required output is 3.

Approach: The problem can be solved using two pointer technique. The idea is to partition the array such that the maximum pair sum of each subset is minimized. Follow the steps below to solve the problem:

• Sort the given array.
• Initialize a variable say, res to store the minimum number of subsets that satisfy the given condition.
• Initialize two variables, say start and end to store the start and end index of the sorted array respectively.
• Traverse the sorted array and check if arr[start] + arr[end] ≤ K or not. If found to be true, then increment the value of start by 1.
• Otherwise, decrement the value of end by 1 and increment the res by 1.
• Finally, print the value of res.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach` `#include ``using` `namespace` `std;` `// Function to get the minimum``// count of subsets that satisfy``// the given condition``int` `cntMinSub(``int` `arr[],``              ``int` `N, ``int` `K)``{``    ``// Store the minimum count``    ``// of subsets that satisfy``    ``// the given condition``    ``int` `res = 0;` `    ``// Stores start index``    ``// of the sorted array.``    ``int` `start = 0;` `    ``// Stores end index``    ``// of the sorted array``    ``int` `end = N - 1;` `    ``// Sort the given array``    ``sort(arr, arr + N);` `    ``// Traverse the array``    ``while` `(end - start > 1) {``        ``if` `(arr[start] + arr[end]``            ``<= K) {``            ``start++;``        ``}``        ``else` `{``            ``res++;``            ``end--;``        ``}``    ``}` `    ``// If only two elements``    ``// of sorted array left``    ``if` `(end - start == 1) {``        ``if` `(arr[start] + arr[end]``            ``<= K) {``            ``res++;``            ``start++;``            ``end--;``        ``}``        ``else` `{``            ``res++;``            ``end--;``        ``}``    ``}` `    ``// If only one elements``    ``// left in the array``    ``if` `(start == end) {``        ``res++;``    ``}` `    ``return` `res;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 2, 6, 8, 10, 20, 25 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);``    ``int` `K = 26;``    ``cout << cntMinSub(arr, N, K);``}`

## Java

 `// Java program to implement``// the above approach``import` `java.util.*;``class` `GFG{` `// Function to get the minimum``// count of subsets that satisfy``// the given condition``static` `int` `cntMinSub(``int` `arr[],``                     ``int` `N, ``int` `K)``{``  ``// Store the minimum count``  ``// of subsets that satisfy``  ``// the given condition``  ``int` `res = ``0``;` `  ``// Stores start index``  ``// of the sorted array.``  ``int` `start = ``0``;` `  ``// Stores end index``  ``// of the sorted array``  ``int` `end = N - ``1``;` `  ``// Sort the given array``  ``Arrays.sort(arr);` `  ``// Traverse the array``  ``while` `(end - start > ``1``)``  ``{``    ``if` `(arr[start] +``        ``arr[end] <= K)``    ``{``      ``start++;``    ``}``    ``else``    ``{``      ``res++;``      ``end--;``    ``}``  ``}` `  ``// If only two elements``  ``// of sorted array left``  ``if` `(end - start == ``1``)``  ``{``    ``if` `(arr[start] +``        ``arr[end] <= K)``    ``{``      ``res++;``      ``start++;``      ``end--;``    ``}``    ``else``    ``{``      ``res++;``      ``end--;``    ``}``  ``}` `  ``// If only one elements``  ``// left in the array``  ``if` `(start == end)``  ``{``    ``res++;``  ``}` `  ``return` `res;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``  ``int` `arr[] = {``2``, ``6``, ``8``, ``10``, ``20``, ``25``};``  ``int` `N = arr.length;``  ``int` `K = ``26``;``  ``System.out.print(cntMinSub(arr, N, K));``}``}` `// This code is contributed by shikhasingrajput`

## Python3

 `# Python3 program to implement``# the above approach` `# Function to get the minimum``# count of subsets that satisfy``# the given condition``def` `cntMinSub(arr, N, K):``    ` `    ``# Store the minimum count``    ``# of subsets that satisfy``    ``# the given condition``    ``res ``=` `0` `    ``# Stores start index``    ``# of the sorted array.``    ``start ``=` `0` `    ``# Stores end index``    ``# of the sorted array``    ``end ``=` `N ``-` `1` `    ``# Sort the given array``    ``arr ``=` `sorted``(arr)` `    ``# Traverse the array``    ``while` `(end ``-` `start > ``1``):``        ``if` `(arr[start] ``+` `arr[end] <``=` `K):``            ``start ``+``=` `1``        ``else``:``            ``res ``+``=` `1``            ``end ``-``=` `1` `    ``# If only two elements``    ``# of sorted array left``    ``if` `(end ``-` `start ``=``=` `1``):``        ``if` `(arr[start] ``+` `arr[end] <``=` `K):``            ``res ``+``=` `1``            ``start ``+``=` `1``            ``end ``-``=` `1``        ``else``:``            ``res ``+``=` `1``            ``end ``-``=` `1``            ` `    ``# If only one elements``    ``# left in the array``    ``if` `(start ``=``=` `end):``        ``res ``+``=` `1` `    ``return` `res` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:` `    ``arr ``=` `[ ``2``, ``6``, ``8``, ``10``, ``20``, ``25` `]``    ``N ``=` `len``(arr)``    ``K ``=` `26``    ` `    ``print``(cntMinSub(arr, N, K))` `# This code is contributed by mohit kumar 29`

## C#

 `// C# program to implement``// the above approach``using` `System;``class` `GFG{` `// Function to get the minimum``// count of subsets that satisfy``// the given condition``static` `int` `cntMinSub(``int` `[]arr,``                     ``int` `N, ``int` `K)``{``  ``// Store the minimum count``  ``// of subsets that satisfy``  ``// the given condition``  ``int` `res = 0;` `  ``// Stores start index``  ``// of the sorted array.``  ``int` `start = 0;` `  ``// Stores end index``  ``// of the sorted array``  ``int` `end = N - 1;` `  ``// Sort the given array``  ``Array.Sort(arr);` `  ``// Traverse the array``  ``while` `(end - start > 1)``  ``{``    ``if` `(arr[start] +``        ``arr[end] <= K)``    ``{``      ``start++;``    ``}``    ``else``    ``{``      ``res++;``      ``end--;``    ``}``  ``}` `  ``// If only two elements``  ``// of sorted array left``  ``if` `(end - start == 1)``  ``{``    ``if` `(arr[start] +``        ``arr[end] <= K)``    ``{``      ``res++;``      ``start++;``      ``end--;``    ``}``    ``else``    ``{``      ``res++;``      ``end--;``    ``}``  ``}` `  ``// If only one elements``  ``// left in the array``  ``if` `(start == end)``  ``{``    ``res++;``  ``}` `  ``return` `res;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``  ``int` `[]arr = {2, 6, 8, 10, 20, 25};``  ``int` `N = arr.Length;``  ``int` `K = 26;``  ``Console.Write(cntMinSub(arr, N, K));``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output

`3`

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

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