Given an array of integers and a number k. We can pair two number of array if difference between them is strictly less than k. The task is to find maximum possible sum of disjoint pairs. Sum of P pairs is sum of all 2P numbers of pairs.

Examples:

Input : arr[] = {3, 5, 10, 15, 17, 12, 9}, K = 4 Output : 62 Then disjoint pairs with difference less than K are, (3, 5), (10, 12), (15, 17) So maximum sum which we can get is 3 + 5 + 12 + 10 + 15 + 17 = 62 Note that an alternate way to form disjoint pairs is, (3, 5), (9, 12), (15, 17), but this pairing produces lesser sum. Input : arr[] = {5, 15, 10, 300}, k = 12 Output : 25

First we sort the given array in increasing order. Once array is sorted, we traverse the array. For every element, we try to pair it with its previous element first. Why do we prefer previous element? Let arr[i] can be paired with arr[i-1] and arr[i-2] (i.e. arr[i] – arr[i-1] < K and arr[i]-arr[i-2] < K). Since the array is sorted, value of arr[i-1] would be more than arr[i-2]. Also, we need to pair with difference less than k, it means if arr[i-2] can be paired, then arr[i-1] can also be paired in a sorted array.

Now observing the above facts, we can formulate our dynamic programming solution as below,

Let dp[i] denotes the maximum disjoint pair sum we can achieve using first i elements of the array. Assume currently we are at i’th position, then there are two possibilities for us.

Pair up i with (i-1)th element, i.e. dp[i] = dp[i-2] + arr[i] + arr[i-1] Don't pair up, i.e. dp[i] = dp[i-1]

Above iteration takes O(N) time and sorting of array will take O(N log N) time so total time complexity of the solution will be O(N log N)

## C++

// C++ program to find maximum pair sum whose // difference is less than K #include <bits/stdc++.h> using namespace std; // method to return maximum sum we can get by // finding less than K difference pair int maxSumPairWithDifferenceLessThanK(int arr[], int N, int K) { // Sort input array in ascending order. sort(arr, arr+N); // dp[i] denotes the maximum disjoint pair sum // we can achieve using first i elements int dp[N]; // if no element then dp value will be 0 dp[0] = 0; for (int i = 1; i < N; i++) { // first give previous value to dp[i] i.e. // no pairing with (i-1)th element dp[i] = dp[i-1]; // if current and previous element can form a pair if (arr[i] - arr[i-1] < K) { // update dp[i] by choosing maximum between // pairing and not pairing if (i >= 2) dp[i] = max(dp[i], dp[i-2] + arr[i] + arr[i-1]); else dp[i] = max(dp[i], arr[i] + arr[i-1]); } } // last index will have the result return dp[N - 1]; } // Driver code to test above methods int main() { int arr[] = {3, 5, 10, 15, 17, 12, 9}; int N = sizeof(arr)/sizeof(int); int K = 4; cout << maxSumPairWithDifferenceLessThanK(arr, N, K); return 0; }

## Python3

# Python3 program to find maximum pair # sum whose difference is less than K # method to return maximum sum we can # get by get by finding less than K # difference pair def maxSumPairWithDifferenceLessThanK(arr, N, K): # Sort input array in ascending order. arr.sort() # dp[i] denotes the maximum disjoint # pair sum we can achieve using first # i elements dp = [0] * N # if no element then dp value will be 0 dp[0] = 0 for i in range(1, N): # first give previous value to # dp[i] i.e. no pairing with # (i-1)th element dp[i] = dp[i-1] # if current and previous element # can form a pair if (arr[i] - arr[i-1] < K): # update dp[i] by choosing # maximum between pairing # and not pairing if (i >= 2): dp[i] = max(dp[i], dp[i-2] + arr[i] + arr[i-1]); else: dp[i] = max(dp[i], arr[i] + arr[i-1]); # last index will have the result return dp[N - 1] # Driver code to test above methods arr = [3, 5, 10, 15, 17, 12, 9] N = len(arr) K = 4 print(maxSumPairWithDifferenceLessThanK(arr, N, K)) # This code is contributed by Smitha Dinesh Semwal

Output:

62

Time complexity : O(N Log N)

Auxiliary Space : O(N)

An optimised solution contributed by Amit Sane is given below,

// C++ program to find maximum pair sum whose // difference is less than K #include <bits/stdc++.h> using namespace std; // Method to return maximum sum we can get by // finding less than K difference pairs int maxSumPairWithDifferenceLessThanK(int arr[], int N, int k) { int maxSum = 0; // Sort elements to ensure every i and i-1 is closest // possible pair sort(arr, arr+N); // To get maximum possible sum, iterate from largest to // smallest, giving larger numbers priority over smaller // numbers. for (int i=N-1; i>0; --i) { // Case I: Diff of arr[i] and arr[i-1] is less then K, // add to maxSum // Case II: Diff between arr[i] and arr[i-1] is not less // then K, move to next i since with sorting we // know, arr[i]-arr[i-1] < arr[i]-arr[i-2] and // so on. if (arr[i]-arr[i-1] < k) { //Assuming only positive numbers. maxSum += arr[i]; maxSum += arr[i-1]; //When a match is found skip this pair --i; } } return maxSum; } // Driver code to test above methods int main() { int arr[] = {3, 5, 10, 15, 17, 12, 9}; int N = sizeof(arr)/sizeof(int); int K = 4; cout << maxSumPairWithDifferenceLessThanK(arr, N, K); return 0; }

Output:

62

Time complexity : O(N Log N)

Auxiliary Space : O(1)

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