Maximum Score from Performing Multiplication Operations

Maximum Score from Performing Multiplication Operations - Problem

You are given two 0-indexed integer arrays nums and multipliers of size n and m respectively, where n >= m.

You begin with a score of 0. You want to perform exactly m operations. On the i-th operation (0-indexed) you will:

Choose one integer x from either the start or the end of the array nums.
Add multipliers[i] * x to your score.
Remove x from nums.

Return the maximum score after performing m operations.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,2,3], multipliers = [3,2,1]

› Output: 14

💡 Note: Optimal picks: Take 3 (right) → score = 3×3 = 9. Take 2 (right) → score = 9 + 2×2 = 13. Take 1 (left) → score = 13 + 1×1 = 14. Total: 14

Example 2 — All From Left

$ Input: nums = [-5,-3,-3,-2,7,1], multipliers = [-10,-5,3,4,6]

› Output: 102

💡 Note: Take from left: -5×(-10)=50, -3×(-5)=15, -3×3=-9, -2×4=-8, 7×6=42. Total: 50+15-9-8+42 = 90. But optimal strategy gives 102.

Example 3 — Mixed Strategy

$ Input: nums = [1], multipliers = [5]

› Output: 5

💡 Note: Only one element and one multiplier: 1×5 = 5

Constraints

n == nums.length
m == multipliers.length
1 ≤ m ≤ 10³
m ≤ n ≤ 10⁵
-1000 ≤ nums[i], multipliers[i] ≤ 1000

Visualization

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Understanding the Visualization

Input Arrays

nums=[1,2,3] and multipliers=[3,2,1]

Pick Strategy

Choose from left or right end, multiply with current multiplier

Maximum Score

Find optimal sequence to maximize total score

Key Takeaway

🎯 Key Insight: Since we can only pick from ends, each state depends on how many we've taken from left vs right, creating perfect DP substructure

Asked in

f Facebook 15 G Google 12 a Amazon 8

The key insight is that we can only pick from the ends of the array, creating overlapping subproblems perfect for dynamic programming. Best approach uses memoization with O(m²) time and space complexity, where each state is defined by (left_picks, operation_number).

Common Approaches

Approach	Time	Space	Notes
✓ Recursive Brute Force	O(2^m)	O(m)	Try all possible combinations of picking from left or right ends
Top-Down DP with Memoization	O(m²)	O(m²)	Cache recursive results to avoid recomputing same subproblems
Bottom-Up 2D DP	O(m²)	O(m²)	Build up solutions iteratively using a 2D DP table

Recursive Brute Force — Algorithm Steps

For each operation, try picking from left end
Try picking from right end
Return maximum of both choices

Visualization

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Step-by-Step Walkthrough

Initial State

Start with full array and first multiplier

Branch Decision

Try picking from both left and right ends

Recursion Tree

Continue branching until all multipliers used

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int max(int a, int b) {
    return a > b ? a : b;
}

int dfs(int* nums, int numsSize, int* multipliers, int multipliersSize, int left, int op) {
    if (op == multipliersSize) {
        return 0;
    }
    
    int right = numsSize - 1 - (op - left);
    
    // Pick from left
    int pickLeft = multipliers[op] * nums[left] + dfs(nums, numsSize, multipliers, multipliersSize, left + 1, op + 1);
    // Pick from right
    int pickRight = multipliers[op] * nums[right] + dfs(nums, numsSize, multipliers, multipliersSize, left, op + 1);
    
    return max(pickLeft, pickRight);
}

int solution(int* nums, int numsSize, int* multipliers, int multipliersSize) {
    return dfs(nums, numsSize, multipliers, multipliersSize, 0, 0);
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000], numsSize = 0;
    char* token = strtok(line + 1, ",]");
    while (token) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    fgets(line, sizeof(line), stdin);
    int multipliers[1000], multipliersSize = 0;
    token = strtok(line + 1, ",]");
    while (token) {
        multipliers[multipliersSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    printf("%d\n", solution(nums, numsSize, multipliers, multipliersSize));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^m)

Each operation has 2 choices, creating 2^m total combinations

✓ Linear Growth

Space Complexity

O(m)

Recursion stack depth equals number of operations m

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Recursive Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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