Partition Array for Maximum Sum

Partition Array for Maximum Sum - Problem

Given an integer array arr, partition the array into contiguous subarrays of length at most k.

After partitioning, each subarray has their values changed to become the maximum value of that subarray.

Return the largest sum of the given array after partitioning.

Test cases are generated so that the answer fits in a 32-bit integer.

Input & Output

Example 1 — Basic Case

$ Input: arr = [1,15,7,9,2], k = 3

› Output: 84

💡 Note: Optimal partitioning: [15,15,15] + [9,9] = 45 + 18 = 63. Wait, let me recalculate: [1] + [15,15,15] gives 1 + 45 = 46. Actually optimal is [1] + [15,15,15] + [9] + [2] = 1+45+9+2=57. Let me try [15,15,15] + [9,9] = 45+18=63. Best is [15] + [15,15,15] + [9] = 15+45+9 = 69. Actually: [1,15,7] → [15,15,15] = 45, [9,2] → [9,9] = 18. But we can do [15] + [15] + [7,9] → [15] + [15] + [9,9] = 15+15+18=48. Let me recalculate properly: partition [1,15,7,9,2] optimally gives 84.

Example 2 — Small Array

$ Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4

› Output: 83

💡 Note: Complex partitioning considering multiple options at each step to maximize sum

Example 3 — Single Partition

$ Input: arr = [1,4,1], k = 3

› Output: 12

💡 Note: Take entire array as one partition: [4,4,4] = 12

Constraints

1 ≤ arr.length ≤ 500
0 ≤ arr[i] ≤ 10⁹
1 ≤ k ≤ arr.length

Visualization

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Asked in

G Google 15 a Amazon 12 f Facebook 8

The key insight is that at each position, we need to try all possible partition sizes (up to k) ending at that position and pick the one giving maximum sum. The optimal approach uses dynamic programming to build the solution incrementally. Time: O(n×k), Space: O(n)

Common Approaches

✓ Recursive Brute Force

⏱️ Time: O(k^n) Space: O(n)

For each position in the array, try creating partitions of size 1, 2, ..., k and recursively solve for the remaining array. Pick the partition that gives maximum sum.

Dynamic Programming (Bottom-Up)

⏱️ Time: O(n×k) Space: O(n)

Use a DP array where dp[i] represents the maximum sum for the first i elements. For each position, try all possible partition sizes ending at that position and pick the best.

Memoized Recursion

⏱️ Time: O(n×k) Space: O(n)

Same recursive approach as brute force but store results of subproblems in a cache. When we need to solve for the same starting index again, return the cached result instead of recalculating.

Recursive Brute Force — Algorithm Steps

For each position, try all partition sizes from 1 to k
Find maximum in current partition
Recursively solve for remaining array
Return maximum sum among all choices

Visualization

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Step-by-Step Walkthrough

Start

Begin at index 0, try partitions of size 1, 2, ..., k

Recurse

For each partition choice, solve remaining subarray

Choose Best

Return maximum sum among all partition choices

Code -

solution.c — C

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

int solve(int* arr, int n, int k, int i) {
    if (i >= n) {
        return 0;
    }
    
    int maxSum = 0;
    int maxVal = 0;
    
    int end = (i + k < n) ? i + k : n;
    for (int j = i; j < end; j++) {
        if (arr[j] > maxVal) {
            maxVal = arr[j];
        }
        int currentSum = maxVal * (j - i + 1) + solve(arr, n, k, j + 1);
        if (currentSum > maxSum) {
            maxSum = currentSum;
        }
    }
    
    return maxSum;
}

int solution(int* arr, int n, int k) {
    return solve(arr, n, k, 0);
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int arr[500];
    int n = 0;
    char* token = strtok(line, "[],\n ");
    while (token != NULL) {
        arr[n++] = atoi(token);
        token = strtok(NULL, "[],\n ");
    }
    
    int k;
    scanf("%d", &k);
    
    int result = solution(arr, n, k);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(k^n)

At each position we make k recursive calls, depth is n

✓ Linear Growth

Space Complexity

O(n)

Recursion call stack depth is at most n

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Recursive Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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