Given an array of integers and a number k. We can pair two numbers of the array if the difference between them is strictly less than k. The task is to find the maximum possible sum of disjoint pairs. Sum of P pairs is the sum of all 2P numbers of pairs.
Examples:
Input : arr[] = {3, 5, 10, 15, 17, 12, 9}, K = 4
Output : 62
Explanation:
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
Approach: 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)
Implementation:
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 pairint 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 methodsint 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;} |
Java
// Java program to find maximum pair sum whose// difference is less than Kimport java.io.*;import java.util.*;class GFG { // method to return maximum sum we can get by // finding less than K difference pair static int maxSumPairWithDifferenceLessThanK(int arr[], int N, int K) { // Sort input array in ascending order. Arrays.sort(arr); // dp[i] denotes the maximum disjoint pair sum // we can achieve using first i elements int dp[] = new int[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] = Math.max(dp[i], dp[i-2] + arr[i] + arr[i-1]); else dp[i] = Math.max(dp[i], arr[i] + arr[i-1]); } } // last index will have the result return dp[N - 1]; } // Driver code to test above methods public static void main (String[] args) { int arr[] = {3, 5, 10, 15, 17, 12, 9}; int N = arr.length; int K = 4; System.out.println ( maxSumPairWithDifferenceLessThanK( arr, N, K)); }}//This code is contributed by vt_m. |
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 pairdef 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 methodsarr = [3, 5, 10, 15, 17, 12, 9]N = len(arr)K = 4print(maxSumPairWithDifferenceLessThanK(arr, N, K))# This code is contributed by Smitha Dinesh Semwal |
C#
// C# program to find maximum pair sum whose// difference is less than Kusing System;class GFG { // method to return maximum sum we can get by // finding less than K difference pair static int maxSumPairWithDifferenceLessThanK(int []arr, int N, int K) { // Sort input array in ascending order. Array.Sort(arr); // dp[i] denotes the maximum disjoint pair sum // we can achieve using first i elements int []dp = new int[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] = Math.Max(dp[i], dp[i-2] + arr[i] + arr[i-1]); else dp[i] = Math.Max(dp[i], arr[i] + arr[i-1]); } } // last index will have the result return dp[N - 1]; } // Driver code to test above methods public static void Main () { int []arr = {3, 5, 10, 15, 17, 12, 9}; int N = arr.Length; int K = 4; Console.WriteLine( maxSumPairWithDifferenceLessThanK(arr, N, K)); }}// This code is contributed by anuj_67. |
PHP
<?php// Php program to find maximum pair sum whose // difference is less than K // method to return maximum sum we can get by // finding less than K difference pair function maxSumPairWithDifferenceLessThanK($arr, $N, $K) { // Sort input array in ascending order. sort($arr) ; // dp[i] denotes the maximum disjoint pair sum // we can achieve using first i elements $dp = array() ; // if no element then dp value will be 0 $dp[0] = 0; for ($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 $arr = array(3, 5, 10, 15, 17, 12, 9); $N = sizeof($arr) ; $K = 4; echo maxSumPairWithDifferenceLessThanK($arr, $N, $K); // This code is contributed by Ryuga?> |
Javascript
<script>// Javascript program to find maximum pair sum whose// difference is less than K // method to return maximum sum we can get by // finding less than K difference pair function 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 let dp = []; // if no element then dp value will be 0 dp[0] = 0; for (let 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] = Math.max(dp[i], dp[i-2] + arr[i] + arr[i-1]); else dp[i] = Math.max(dp[i], arr[i] + arr[i-1]); } } // last index will have the result return dp[N - 1]; }// Driver code to test above methods let arr = [3, 5, 10, 15, 17, 12, 9]; let N = arr.length; let K = 4; document.write( maxSumPairWithDifferenceLessThanK( arr, N, K));// This code is contributed by avijitmondal1998.</script> |
Output
62
Time complexity: O(N Log N)
Auxiliary Space: O(N)
An optimised solution contributed by Amit Sane is given below,
Implementation:
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 pairsint maxSumPair(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 than K,add to maxSum // Case II: Diff between arr[i] and arr[i-1] is not // less than K, move to next i since with // sorting we know, arr[i]-arr[i-1] < // rr[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 codeint main(){ int arr[] = { 3, 5, 10, 15, 17, 12, 9 }; int N = sizeof(arr) / sizeof(int); int K = 4; cout << maxSumPair(arr, N, K); return 0;} |
Java
// Java program to find maximum pair sum whose// difference is less than Kimport java.io.*;import java.util.*;class GFG { // Method to return maximum sum we can get by // finding less than K difference pairs static int maxSumPairWithDifferenceLessThanK(int arr[], int N, int k) { int maxSum = 0; // Sort elements to ensure every i and i-1 is // closest possible pair Arrays.sort(arr); // 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 // than K, add to maxSum // Case II: Diff between arr[i] and arr[i-1] is // not less than 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 public static void main(String[] args) { int arr[] = { 3, 5, 10, 15, 17, 12, 9 }; int N = arr.length; int K = 4; System.out.println( maxSumPairWithDifferenceLessThanK(arr, N, K)); }}// This code is contributed by vt_m. |
Python3
# Python3 program to find maximum pair sum# whose difference is less than K# Method to return maximum sum we can# get by finding less than K difference# pairsdef maxSumPairWithDifferenceLessThanK(arr, N, k): maxSum = 0 # Sort elements to ensure every i and # i-1 is closest possible pair arr.sort() # To get maximum possible sum, iterate # from largest to smallest, giving larger # numbers priority over smaller numbers. i = N - 1 while (i > 0): # Case I: Diff of arr[i] and arr[i-1] # is less than K, add to maxSum # Case II: Diff between arr[i] and # arr[i-1] is not less than 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 -= 1 i -= 1 return maxSum# Driver Codearr = [3, 5, 10, 15, 17, 12, 9]N = len(arr)K = 4print(maxSumPairWithDifferenceLessThanK(arr, N, K))# This code is contributed by mits |
C#
// C# program to find maximum pair sum whose// difference is less than Kusing System;class GFG { // Method to return maximum sum we can get by // finding less than K difference pairs static int maxSumPairWithDifferenceLessThanK(int []arr, int N, int k) { int maxSum = 0; // Sort elements to ensure // every i and i-1 is closest // possible pair Array.Sort(arr); // 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 than K, add to maxSum Case II: Diff between arr[i] and arr[i-1] is not less than 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 public static void Main () { int []arr = {3, 5, 10, 15, 17, 12, 9}; int N = arr.Length; int K = 4; Console.Write( maxSumPairWithDifferenceLessThanK(arr, N, K)); }}// This code is contributed by nitin mittal. |
PHP
<?php// PHP program to find maximum pair sum // whose difference is less than K // Method to return maximum sum we can // get by finding less than K difference// pairs function maxSumPairWithDifferenceLessThanK($arr, $N, $k) { $maxSum = 0; // Sort elements to ensure every i and // i-1 is closest possible pair sort($arr); // To get maximum possible sum, iterate // from largest to smallest, giving larger // numbers priority over smaller numbers. for ($i = $N - 1; $i > 0; --$i) { // Case I: Diff of arr[i] and arr[i-1] // is less than K, add to maxSum // Case II: Diff between arr[i] and // arr[i-1] is not less than 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 $maxSum; } // Driver Code$arr = array(3, 5, 10, 15, 17, 12, 9); $N = sizeof($arr); $K = 4; echo maxSumPairWithDifferenceLessThanK($arr, $N, $K); // This code is contributed // by Sach_Code ?> |
Javascript
<script>// Javascript program to find// maximum pair sum whose// difference is less than K// Method to return maximum sum we can get by// finding less than K difference pairsfunction maxSumPairWithDifferenceLessThanK(arr, N, k){ var maxSum = 0; // Sort elements to ensure every i and i-1 is // closest possible pair arr.sort((a,b)=>a-b); // To get maximum possible sum, // iterate from largest // to smallest, giving larger // numbers priority over // smaller numbers. for (i = N - 1; i > 0; --i) { // Case I: Diff of arr[i] and arr[i-1] is less // than K, add to maxSum // Case II: Diff between arr[i] and arr[i-1] is // not less than 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 codevar arr = [ 3, 5, 10, 15, 17, 12, 9 ];var N = arr.length;var K = 4;document.write(maxSumPairWithDifferenceLessThanK(arr, N, K));// This code is contributed by 29AjayKumar </script> |
Output
62
Time complexity: O(N Log N)
Auxiliary Space: O(1)
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