LeetCode #3836 — HARD

Maximum Score Using Exactly K Pairs

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

You are given two integer arrays nums1 and nums2 of lengths n and m respectively, and an integer k.

You must choose exactly k pairs of indices (i₁, j₁), (i₂, j₂), ..., (i_k, j_k) such that:

0 <= i₁ < i₂ < ... < i_k < n
0 <= j₁ < j₂ < ... < j_k < m

For each chosen pair (i, j), you gain a score of nums1[i] * nums2[j].

The total score is the sum of the products of all selected pairs.

Return an integer representing the maximum achievable total score.

Example 1:

Input: nums1 = [1,3,2], nums2 = [4,5,1], k = 2

Output: 22

Explanation:

One optimal choice of index pairs is:

(i₁, j₁) = (1, 0) which scores 3 * 4 = 12
(i₂, j₂) = (2, 1) which scores 2 * 5 = 10

This gives a total score of 12 + 10 = 22.

Example 2:

Input: nums1 = [-2,0,5], nums2 = [-3,4,-1,2], k = 2

Output: 26

Explanation:

One optimal choice of index pairs is:

(i₁, j₁) = (0, 0) which scores -2 * -3 = 6
(i₂, j₂) = (2, 1) which scores 5 * 4 = 20

The total score is 6 + 20 = 26.

Example 3:

Input: nums1 = [-3,-2], nums2 = [1,2], k = 2

Output: -7

Explanation:

The optimal choice of index pairs is:

(i₁, j₁) = (0, 0) which scores -3 * 1 = -3
(i₂, j₂) = (1, 1) which scores -2 * 2 = -4

The total score is -3 + (-4) = -7.

Constraints:

1 <= n == nums1.length <= 100
1 <= m == nums2.length <= 100
-10⁶ <= nums1[i], nums2[i] <= 10⁶
1 <= k <= min(n, m)

Step 01

Brute Force Baseline

Problem summary: You are given two integer arrays nums1 and nums2 of lengths n and m respectively, and an integer k. You must choose exactly k pairs of indices (i1, j1), (i2, j2), ..., (ik, jk) such that: 0 <= i1 < i2 < ... < ik < n 0 <= j1 < j2 < ... < jk < m For each chosen pair (i, j), you gain a score of nums1[i] * nums2[j]. The total score is the sum of the products of all selected pairs. Return an integer representing the maximum achievable total score.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

[1,3,2]
[4,5,1]
2

Example 2

[-2,0,5]
[-3,4,-1,2]
2

Example 3

[-3,-2]
[1,2]
2

Step 02

Core Insight

What unlocks the optimal approach

Use dynamic programming
Let dynamic programming states <code>dp(i, j, k)</code>, denote the maximum <code>k-th</code> pair sum you can get from the first <code>nums1[0..i]</code> and <code>nums2[0..j]</code>
Transition: <code>dp(i, j, t) = max(dp(i - 1, j, t), dp(i, j - 1, t), dp(i - 1, j - 1, t - 1) + nums1[i] * nums2[j])</code>

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3836: Maximum Score Using Exactly K Pairs
class Solution {
    public long maxScore(int[] nums1, int[] nums2, int K) {
        int n = nums1.length, m = nums2.length;
        long NEG = Long.MIN_VALUE / 4;
        long[][][] f = new long[n + 1][m + 1][K + 1];
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= m; j++) {
                Arrays.fill(f[i][j], NEG);
            }
        }
        f[0][0][0] = 0;
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= m; j++) {
                for (int k = 0; k <= K; k++) {
                    if (i > 0) {
                        f[i][j][k] = Math.max(f[i][j][k], f[i - 1][j][k]);
                    }
                    if (j > 0) {
                        f[i][j][k] = Math.max(f[i][j][k], f[i][j - 1][k]);
                    }
                    if (i > 0 && j > 0 && k > 0) {
                        f[i][j][k] = Math.max(f[i][j][k],
                            f[i - 1][j - 1][k - 1] + (long) nums1[i - 1] * nums2[j - 1]);
                    }
                }
            }
        }
        return f[n][m][K];
    }
}

// Accepted solution for LeetCode #3836: Maximum Score Using Exactly K Pairs
func maxScore(nums1 []int, nums2 []int, K int) int64 {
	n, m := len(nums1), len(nums2)
	NEG := int64(math.MinInt64 / 4)
	f := make([][][]int64, n+1)
	for i := 0; i <= n; i++ {
		f[i] = make([][]int64, m+1)
		for j := 0; j <= m; j++ {
			f[i][j] = make([]int64, K+1)
			for k := 0; k <= K; k++ {
				f[i][j][k] = NEG
			}
		}
	}
	f[0][0][0] = 0
	for i := 0; i <= n; i++ {
		for j := 0; j <= m; j++ {
			for k := 0; k <= K; k++ {
				if i > 0 {
					f[i][j][k] = max(f[i][j][k], f[i-1][j][k])
				}
				if j > 0 {
					f[i][j][k] = max(f[i][j][k], f[i][j-1][k])
				}
				if i > 0 && j > 0 && k > 0 {
					f[i][j][k] = max(
						f[i][j][k],
						f[i-1][j-1][k-1]+int64(nums1[i-1])*int64(nums2[j-1]),
					)
				}
			}
		}
	}
	return f[n][m][K]
}

# Accepted solution for LeetCode #3836: Maximum Score Using Exactly K Pairs
class Solution:
    def maxScore(self, nums1: List[int], nums2: List[int], K: int) -> int:
        n, m = len(nums1), len(nums2)
        f = [[[-inf] * (K + 1) for _ in range(m + 1)] for _ in range(n + 1)]
        f[0][0][0] = 0
        for i in range(n + 1):
            for j in range(m + 1):
                for k in range(K + 1):
                    if i > 0:
                        f[i][j][k] = max(f[i][j][k], f[i - 1][j][k])
                    if j > 0:
                        f[i][j][k] = max(f[i][j][k], f[i][j - 1][k])
                    if i > 0 and j > 0 and k > 0:
                        f[i][j][k] = max(
                            f[i][j][k],
                            f[i - 1][j - 1][k - 1] + nums1[i - 1] * nums2[j - 1],
                        )
        return f[n][m][K]

// Accepted solution for LeetCode #3836: Maximum Score Using Exactly K Pairs
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3836: Maximum Score Using Exactly K Pairs
// class Solution {
//     public long maxScore(int[] nums1, int[] nums2, int K) {
//         int n = nums1.length, m = nums2.length;
//         long NEG = Long.MIN_VALUE / 4;
//         long[][][] f = new long[n + 1][m + 1][K + 1];
//         for (int i = 0; i <= n; i++) {
//             for (int j = 0; j <= m; j++) {
//                 Arrays.fill(f[i][j], NEG);
//             }
//         }
//         f[0][0][0] = 0;
//         for (int i = 0; i <= n; i++) {
//             for (int j = 0; j <= m; j++) {
//                 for (int k = 0; k <= K; k++) {
//                     if (i > 0) {
//                         f[i][j][k] = Math.max(f[i][j][k], f[i - 1][j][k]);
//                     }
//                     if (j > 0) {
//                         f[i][j][k] = Math.max(f[i][j][k], f[i][j - 1][k]);
//                     }
//                     if (i > 0 && j > 0 && k > 0) {
//                         f[i][j][k] = Math.max(f[i][j][k],
//                             f[i - 1][j - 1][k - 1] + (long) nums1[i - 1] * nums2[j - 1]);
//                     }
//                 }
//             }
//         }
//         return f[n][m][K];
//     }
// }

// Accepted solution for LeetCode #3836: Maximum Score Using Exactly K Pairs
function maxScore(nums1: number[], nums2: number[], K: number): number {
    const n = nums1.length,
        m = nums2.length;
    const NEG = -1e18;
    const f = Array.from({ length: n + 1 }, () =>
        Array.from({ length: m + 1 }, () => Array(K + 1).fill(NEG)),
    );
    f[0][0][0] = 0;
    for (let i = 0; i <= n; i++) {
        for (let j = 0; j <= m; j++) {
            for (let k = 0; k <= K; k++) {
                if (i > 0) {
                    f[i][j][k] = Math.max(f[i][j][k], f[i - 1][j][k]);
                }
                if (j > 0) {
                    f[i][j][k] = Math.max(f[i][j][k], f[i][j - 1][k]);
                }
                if (i > 0 && j > 0 && k > 0) {
                    f[i][j][k] = Math.max(
                        f[i][j][k],
                        f[i - 1][j - 1][k - 1] + nums1[i - 1] * nums2[j - 1],
                    );
                }
            }
        }
    }
    return f[n][m][K];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n × K)

Space

O(m × n × K)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.