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K maximum sum combinations from two arrays

Last Updated : 02 Jul, 2025
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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.

Table of Content

[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;
}
Java 
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 + " ");
        }
    }
}
Python 
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=" ")
C# 
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 + " ");
        }
    }
}
JavaScript 
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(" "));

Output
10 9 9

[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;
}
Java 
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 + " ");
        }
    }
}
Python 
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)))
C# 
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 + " ");
        }
    }
}
JavaScript 
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(" "));

Output
7 6

[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
Java 
//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
Python 
#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
C# 
//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
JavaScript 
// 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

Output
10 9 9



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