Partition Array for Maximum Sum - Practice Coding Problems

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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Understanding the Visualization

Input

Array [1,15,7,9,2] with k=3 (max partition size)

Partition

Find optimal way to split into subarrays of size ≤ k

Transform

Replace each partition with its maximum value

Sum

Calculate total sum of transformed array

Key Takeaway

🎯 Key Insight: Try all possible partition sizes at each position and use dynamic programming to find the optimal solution

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

Approach	Time	Space	Notes
✓ Recursive Brute Force	O(k^n)	O(n)	Try all possible partitions recursively
Memoized Recursion	O(n×k)	O(n)	Cache recursive results to avoid recalculation
Dynamic Programming (Bottom-Up)	O(n×k)	O(n)	Build solution iteratively from left to right

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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Smart Actions

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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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