K maximum sum combinations from two arrays
Last Updated :
02 Jul, 2025
Given two integer arrays a[] and b[] of the same length, and an positive integer k, the goal is to find the top k maximum sum combinations, where each combination is formed by adding one element from a and one from b.
Each index from both arrays can be used at most once in a pair. Return the k largest sums in descending order.
Examples:
Input: a[] = [3, 2], b[] = [1, 4], k = 2
Output: [7, 6]
Explanation: Possible sums: 3 + 1 = 4, 3 + 4 = 7, 2 + 1 = 3, 2 + 4 = 6, Top 2 sums are 7 and 6.
Input: a[] = [1, 4, 2, 3], b[] = [2, 5, 1, 6], k = 3
Output: [10, 9, 9]
Explanation: The top 3 maximum possible sums are : 4 + 6 = 10, 3 + 6 = 9, and 4 + 5 = 9.
[Naive Approach] Generate All Combinations - O(n2 × log (n2)) Time and O(n2) Space
The idea is to compute all possible sum combinations by pairing each element from array a[] with every element from array b[]. Since each element in a can form a pair with every element in b, we use a nested loop to generate all such sums. After generating these sums, we sort them in descending order and collect the top k values as the final result.
C++
#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
vector<int> topKSumPairs(vector<int>& a, vector<int>& b, int k) {
int n = a.size();
// store all possible pair sums
vector<int> allpossible;
// generate all possible sums by pairing every
// element of 'a' with every element of 'b'
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
int currSum = a[i] + b[j];
allpossible.push_back(currSum);
}
}
// sort the sums in descending order to
// get the largest ones first
sort(allpossible.begin(), allpossible.end(), greater<int>());
// extract the top 'k' largest sums
vector<int> res;
for (int i = 0; i < k; i++) {
res.push_back(allpossible[i]);
}
return res;
}
int main() {
vector<int> a = {1, 4, 2, 3};
vector<int> b = {2, 5, 1, 6};
int k = 3;
vector<int> res = topKSumPairs(a, b, k);
for (int sum : res) {
cout << sum << " ";
}
return 0;
}
import java.util.ArrayList;
import java.util.Collections;
class GfG {
public static ArrayList<Integer> topKSumPairs(int[] a, int[] b, int k) {
int n = a.length;
// store all possible pair sums
ArrayList<Integer> allpossible = new ArrayList<>();
// generate all possible sums by pairing every
// element of 'a' with every element of 'b'
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
int currSum = a[i] + b[j];
allpossible.add(currSum);
}
}
// sort the sums in descending order to
// get the largest ones first
Collections.sort(allpossible, Collections.reverseOrder());
// extract the top 'k' largest sums
ArrayList<Integer> res = new ArrayList<>();
for (int i = 0; i < k; i++) {
res.add(allpossible.get(i));
}
return res;
}
public static void main(String[] args) {
int[] a = {1, 4, 2, 3};
int[] b = {2, 5, 1, 6};
int k = 3;
ArrayList<Integer> res = topKSumPairs(a, b, k);
for (int sum : res) {
System.out.print(sum + " ");
}
}
}
def topKSumPairs(a, b, k):
n = len(a)
# store all possible pair sums
allpossible = []
# generate all possible sums by pairing every
# element of 'a' with every element of 'b'
for i in range(n):
for j in range(n):
currSum = a[i] + b[j]
allpossible.append(currSum)
# sort the sums in descending order to
# get the largest ones first
allpossible.sort(reverse=True)
# extract the top 'k' largest sums
res = []
for i in range(k):
res.append(allpossible[i])
return res
if __name__ == "__main__":
a = [1, 4, 2, 3]
b = [2, 5, 1, 6]
k = 3
res = topKSumPairs(a, b, k)
for sum in res:
print(sum, end=" ")
using System;
using System.Collections.Generic;
class GfG {
public static List<int> topKSumPairs(int[] a, int[] b, int k) {
int n = a.Length;
// store all possible pair sums
List<int> allpossible = new List<int>();
// generate all possible sums by pairing every
// element of 'a' with every element of 'b'
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
int currSum = a[i] + b[j];
allpossible.Add(currSum);
}
}
// sort the sums in descending order to
// get the largest ones first
allpossible.Sort((x, y) => y.CompareTo(x));
// extract the top 'k' largest sums
List<int> res = new List<int>();
for (int i = 0; i < k; i++) {
res.Add(allpossible[i]);
}
return res;
}
public static void Main() {
int[] a = {1, 4, 2, 3};
int[] b = {2, 5, 1, 6};
int k = 3;
List<int> res = topKSumPairs(a, b, k);
foreach (int sum in res) {
Console.Write(sum + " ");
}
}
}
function topKSumPairs(a, b, k) {
const n = a.length;
// store all possible pair sums
let allpossible = [];
// generate all possible sums by pairing every
// element of 'a' with every element of 'b'
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++) {
let currSum = a[i] + b[j];
allpossible.push(currSum);
}
}
// sort the sums in descending order to
// get the largest ones first
allpossible.sort((x, y) => y - x);
// extract the top 'k' largest sums
let res = [];
for (let i = 0; i < k; i++) {
res.push(allpossible[i]);
}
return res;
}
// Driver Code
let a = [ 1, 4, 2, 3 ];
let b = [ 2, 5, 1, 6 ];
let k = 3;
let res = topKSumPairs(a, b, k);
console.log(res.join(" "));
[Better Approach] Heap-Based Top-K Pair Sum Approach - O(n2 × log k) Time and O(k) Space
The idea is to generate all possible pair sums by adding each element from array a[] to each element from array b[]. A min-heap of size k is used to maintain the top k maximum sums. For every new pair sum, we insert it into the heap if it’s among the largest k seen so far. Finally, we extract and reverse the heap to return the top k sums in descending order.
C++
#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
vector<int> topKSumPairs(vector<int>& a, vector<int>& b, int k) {
int n = a.size();
// min-heap to keep track of the top k maximum pair sums
priority_queue<int, vector<int>, greater<int>> minHeap;
// generate all possible pair sums from a[i] + b[j]
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
int currSum = a[i] + b[j];
// the heap has less than k elements, add the current sum
if ((int)minHeap.size() < k) {
minHeap.push(currSum);
}
// current sum is greater than the smallest in heap, replace it
else if (currSum > minHeap.top()) {
minHeap.pop();
minHeap.push(currSum);
}
}
}
// extract elements from the heap and store in result
vector<int> res;
while (!minHeap.empty()) {
res.push_back(minHeap.top());
minHeap.pop();
}
// convert to descending order as required
reverse(res.begin(), res.end());
return res;
}
int main() {
vector<int> a = {1, 4, 2, 3};
vector<int> b = {2, 5, 1, 6};
int k = 3;
vector<int> res = topKSumPairs(a, b, k);
for (int sum : res) {
cout << sum << " ";
}
return 0;
}
import java.util.PriorityQueue;
import java.util.Collections;
import java.util.ArrayList;
class GfG {
public static ArrayList<Integer> topKSumPairs(int[] a, int[] b, int k) {
int n = a.length;
// min-heap to keep track of the top k maximum pair sums
PriorityQueue<Integer> minHeap = new PriorityQueue<>();
// generate all possible pair sums from a[i] + b[j]
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
int currSum = a[i] + b[j];
// the heap has less than k elements, add the current sum
if (minHeap.size() < k) {
minHeap.offer(currSum);
}
// current sum is greater than the smallest in heap, replace it
else if (currSum > minHeap.peek()) {
minHeap.poll();
minHeap.offer(currSum);
}
}
}
// Extract elements from the heap and store in result
ArrayList<Integer> res = new ArrayList<>();
while (!minHeap.isEmpty()) {
res.add(minHeap.poll());
}
// Convert to descending order as required
Collections.reverse(res);
return res;
}
public static void main(String[] args) {
int[] a = {1, 4, 2, 3};
int[] b = {2, 5, 1, 6};
int k = 3;
ArrayList<Integer> res = topKSumPairs(a, b, k);
for (int sum : res) {
System.out.print(sum + " ");
}
}
}
import heapq
def topKSumPairs(a, b, k):
n = len(a)
# min-heap to keep track of the top k maximum pair sums
minHeap = []
# generate all possible pair sums from a[i] + b[j]
for i in range(n):
for j in range(n):
currSum = a[i] + b[j]
# the heap has less than k elements, add the current sum
if len(minHeap) < k:
heapq.heappush(minHeap, currSum)
# current sum is greater than the smallest pair
# sum in the heap then replace it
elif currSum > minHeap[0]:
heapq.heappop(minHeap)
heapq.heappush(minHeap, currSum)
# extract elements from heap and store in result
res = []
while minHeap:
res.append(heapq.heappop(minHeap))
# convert to descending order as required
res.reverse()
return res
if __name__ == "__main__":
a = [1, 4, 2, 3]
b = [2, 5, 1, 6]
k = 3
res = topKSumPairs(a, b, k)
print(" ".join(map(str, res)))
using System;
using System.Collections.Generic;
class GfG {
public static List<int> topKSumPairs(int[] a, int[] b, int k) {
int n = a.Length;
// Min-heap to keep track of top k largest pair sums
SortedSet<(int sum, int index)> minHeap = new SortedSet<(int, int)>();
int idx = 0;
// Generate all possible pair sums
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
int currSum = a[i] + b[j];
if (minHeap.Count < k) {
minHeap.Add((currSum, idx++));
}
else {
// If current sum is greater than the smallest in heap
if (currSum > minHeap.Min.sum) {
minHeap.Remove(minHeap.Min);
minHeap.Add((currSum, idx++));
}
}
}
}
// Extract results and sort in descending order
List<int> res = new List<int>();
foreach (var pair in minHeap) {
res.Add(pair.sum);
}
// Descending order
res.Sort((x, y) => y.CompareTo(x));
return res;
}
public static void Main() {
int[] a = {1, 4, 2, 3};
int[] b = {2, 5, 1, 6};
int k = 3;
List<int> res = topKSumPairs(a, b, k);
foreach (int sum in res) {
Console.Write(sum + " ");
}
}
}
function topKSumPairs(a, b, k) {
let n = a.length;
// min-heap to keep track of the top k maximum pair sums
let minHeap = [];
// generate all possible pair sums from a[i] + b[j]
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++) {
let currSum = a[i] + b[j];
// the heap has less than k elements, add the current sum
if (minHeap.length < k) {
minHeap.push(currSum);
minHeap.sort((x, y) => x - y);
}
// current sum is greater than the smallest pair
// sum in the heap then replace it
else if (currSum > minHeap[0]) {
minHeap.shift(); // remove smallest
minHeap.push(currSum);
minHeap.sort((x, y) => x - y);
}
}
}
// convert to descending order as required
return minHeap.sort((a, b) => b - a);
}
// Driver Code
const a = [1, 4, 2, 3];
const b = [2, 5, 1, 6];
const k = 3;
const res = topKSumPairs(a, b, k);
console.log(res.join(" "));
[Expected Approach] Max Heap with Index Tracking - O(n × log n) Time and O(k) Space
The idea is to combine the largest elements from both arrays to form the highest possible sums. By sorting the arrays in descending order, we ensure that the sum at indices (0, 0) is the maximum. From this point, the next potential largest sums lie at positions (i+1, j) and (i, j+1) because of the sorted order. By exploring these candidates using a max heap and tracking visited index pairs, we efficiently gather the top k combinations without redundantly computing all n2 possibilities.
Steps by step approach:
- Sort both arrays
a[] and b[] in descending order to ensure the largest sums are considered first. - Initialize a max heap and insert the sum of the first elements
(a[0] + b[0]) along with their indices (0, 0). - Use a set to track visited index pairs to prevent reprocessing the same combinations.
- Repeat the following for
k iterations to extract the top k sums:- Pop the current largest sum from the heap and add it to the result.
- Push the next two possible combinations:
(i+1, j) and (i, j+1) into the heap if within bounds and not visited.
C++
//Driver Code Starts
#include <iostream>
#include <vector>
#include <queue>
#include <set>
#include <algorithm>
using namespace std;
//Driver Code Ends
vector<int> topKSumPairs(vector<int> &a, vector<int> &b, int k) {
int n = a.size();
// sort both arrays in descending order
sort(a.rbegin(), a.rend());
sort(b.rbegin(), b.rend());
// max-heap to store {sum, {i, j}}
priority_queue<pair<int, pair<int, int>>> pq;
set<pair<int, int>> vis;
// start from the largest possible pair (0,0)
pq.push({a[0] + b[0], {0, 0}});
vis.insert({0, 0});
vector<int> res;
while (res.size() < k) {
auto top = pq.top();
pq.pop();
int sum = top.first;
int i = top.second.first;
int j = top.second.second;
res.push_back(sum);
// add next element in a if not visited
if (i + 1 < n && vis.find({i + 1, j}) == vis.end()) {
pq.push({a[i + 1] + b[j], {i + 1, j}});
vis.insert({i + 1, j});
}
// add next element in b if not visited
if (j + 1 < n && vis.find({i, j + 1}) == vis.end()) {
pq.push({a[i] + b[j + 1], {i, j + 1}});
vis.insert({i, j + 1});
}
}
//Driver Code Starts
return res;
}
int main() {
vector<int> a = {1, 4, 2, 3};
vector<int> b = {2, 5, 1, 6};
int k = 3;
vector<int> result = topKSumPairs(a, b, k);
for (int sum : result) {
cout << sum << " ";
}
return 0;
}
//Driver Code Ends
//Driver Code Starts
import java.util.ArrayList;
import java.util.Arrays;
import java.util.PriorityQueue;
//Driver Code Ends
import java.util.HashSet;
class GfG {
public static ArrayList<Integer> topKSumPairs(int[] a, int[] b, int k) {
int n = a.length;
// Sort both arrays in descending order
Arrays.sort(a);
Arrays.sort(b);
reverse(a);
reverse(b);
// Max-heap to store {sum, i, j}
PriorityQueue<int[]> pq = new PriorityQueue<>((x, y) -> y[0] - x[0]);
HashSet<String> vis = new HashSet<>();
// Start from the largest possible pair (0, 0)
pq.offer(new int[]{a[0] + b[0], 0, 0});
vis.add("0#0");
ArrayList<Integer> res = new ArrayList<>();
while (res.size() < k && !pq.isEmpty()) {
int[] top = pq.poll();
int sum = top[0], i = top[1], j = top[2];
res.add(sum);
// Try the next element in array a
if (i + 1 < n && !vis.contains((i + 1) + "#" + j)) {
pq.offer(new int[]{a[i + 1] + b[j], i + 1, j});
vis.add((i + 1) + "#" + j);
}
// Try the next element in array b
if (j + 1 < n && !vis.contains(i + "#" + (j + 1))) {
pq.offer(new int[]{a[i] + b[j + 1], i, j + 1});
vis.add(i + "#" + (j + 1));
}
}
//Driver Code Starts
return res;
}
// Helper method to reverse an array in-place
private static void reverse(int[] arr) {
int l = 0, r = arr.length - 1;
while (l < r) {
int temp = arr[l];
arr[l++] = arr[r];
arr[r--] = temp;
}
}
public static void main(String[] args) {
int[] a = {1, 4, 2, 3};
int[] b = {2, 5, 1, 6};
int k = 3;
ArrayList<Integer> result = topKSumPairs(a, b, k);
for (int val : result) {
System.out.print(val + " ");
}
}
}
//Driver Code Ends
#Driver Code Starts
import heapq
#Driver Code Ends
def topKSumPairs(a, b, k):
n = len(a)
# sort both arrays in descending order
a.sort(reverse=True)
b.sort(reverse=True)
visited = set()
heap = []
# max-heap to store {sum, {i, j}}
heapq.heappush(heap, (-(a[0] + b[0]), 0, 0))
# start from the largest possible pair (0,0)
visited.add((0, 0))
res = []
while len(res) < k:
sum_neg, i, j = heapq.heappop(heap)
res.append(-sum_neg)
# add next element in a if not visited
if i + 1 < n and (i + 1, j) not in visited:
heapq.heappush(heap, (-(a[i + 1] + b[j]), i + 1, j))
visited.add((i + 1, j))
# add next element in b if not visited
if j + 1 < n and (i, j + 1) not in visited:
heapq.heappush(heap, (-(a[i] + b[j + 1]), i, j + 1))
visited.add((i, j + 1))
return res
#Driver Code Starts
if __name__ == "__main__":
a = [1, 4, 2, 3]
b = [2, 5, 1, 6]
k = 3
result = topKSumPairs(a, b, k)
print(" ".join(map(str, result)))
#Driver Code Ends
//Driver Code Starts
using System;
using System.Collections.Generic;
class GfG {
//Driver Code Ends
public static List<int> topKSumPairs(int[] a, int[] b, int k) {
int n = a.Length;
// Sort both arrays in descending order
Array.Sort(a);
Array.Sort(b);
Array.Reverse(a);
Array.Reverse(b);
// Max-heap: SortedSet stores (sum, i, j) sorted by descending sum
var maxHeap = new SortedSet<(int, int, int)>(
Comparer<(int, int, int)>.Create((x, y) =>
x.Item1 != y.Item1 ? y.Item1.CompareTo(x.Item1) :
x.Item2 != y.Item2 ? x.Item2.CompareTo(y.Item2) :
x.Item3.CompareTo(y.Item3)
)
);
var visited = new HashSet<(int, int)>();
var result = new List<int>();
// Add initial pair (0, 0)
maxHeap.Add((a[0] + b[0], 0, 0));
visited.Add((0, 0));
while (result.Count < k && maxHeap.Count > 0) {
var top = maxHeap.Min;
maxHeap.Remove(top);
int sum = top.Item1;
int i = top.Item2, j = top.Item3;
result.Add(sum);
// Move to next element in a
if (i + 1 < n && !visited.Contains((i + 1, j))) {
maxHeap.Add((a[i + 1] + b[j], i + 1, j));
visited.Add((i + 1, j));
//Driver Code Starts
}
// Move to next element in b
if (j + 1 < n && !visited.Contains((i, j + 1))) {
maxHeap.Add((a[i] + b[j + 1], i, j + 1));
visited.Add((i, j + 1));
}
}
return result;
}
public static void Main() {
int[] a = {1, 4, 2, 3};
int[] b = {2, 5, 1, 6};
int k = 3;
List<int> res = topKSumPairs(a, b, k);
foreach (int sum in res) {
Console.Write(sum + " ");
}
}
}
//Driver Code Ends
// Custom Max Heap implementation
class MaxHeap {
constructor() {
this.heap = [];
}
push(element) {
this.heap.push(element);
// Sort the heap to maintain max-heap property (descending order)
this.heap.sort((a, b) => b[0] - a[0]);
}
pop() {
return this.heap.shift();
}
size() {
return this.heap.length;
}
}
// Function to find top k maximum sum combinations
function topKSumPairs(a, b, k) {
let n = a.length;
// Sort both arrays in descending order
a.sort((x, y) => y - x);
b.sort((x, y) => y - x);
const heap = new MaxHeap();
const used = new Set();
// Start from the largest pair (0, 0)
heap.push([a[0] + b[0], 0, 0]);
used.add(`0,0`);
const res = [];
//Driver Code Starts
// Extract top k combinations
while (res.length < k) {
const [sum, i, j] = heap.pop();
res.push(sum);
// Next element from a
if (i + 1 < n && !used.has(`${i + 1},${j}`)) {
heap.push([a[i + 1] + b[j], i + 1, j]);
used.add(`${i + 1},${j}`);
}
// Next element from b
if (j + 1 < n && !used.has(`${i},${j + 1}`)) {
heap.push([a[i] + b[j + 1], i, j + 1]);
used.add(`${i},${j + 1}`);
}
}
return res;
}
// Driver Code
const k = 3;
const a = [1, 4, 2, 3];
const b = [2, 5, 1, 6];
const result = topKSumPairs(a, b, k);
console.log(result.join(" "));
//Driver Code Ends
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Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
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Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
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Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
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Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
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Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
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Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
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Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
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GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
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