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Leetcode 3046. Split the Array

Problem Statement

You are given an integer array nums and an integer k. Split the array into k non-empty subarrays such that the sum of the subarrays’ widths is minimized.

The width of an array is defined as the difference between the maximum and minimum elements in the array.

Clarifying Questions

  1. Range of nums and k: What are the constraints on the size of the nums array and the value of k?
    • Typical Response: Suppose nums can have up to 1000 elements, and k is a positive integer such that 1 <= k <= nums.size().
  2. Negative Numbers: Can nums contain negative numbers?
    • Typical Response: Yes, nums can contain negative as well as positive numbers.
  3. Sum Minimization: We’re minimizing the sum of the widths of k subarrays. The width is defined as max(subarray) - min(subarray)?
    • Typical Response: Correct.

Strategy

To solve this problem we can use dynamic programming to handle the subarray calculations efficiently. Here’s a general approach:

  1. Sorting: Sort the array nums to help in quickly finding the min and max in each subarray.
  2. DP Table: Define a DP table dp[i][j] where i represents the first i elements of the array and j represents j subarrays.
  3. Transition: Use nested loops to fill the DP table by considering the cost of making a new subarray at different partition points.

Time Complexity

The time complexity for this approach would typically be O(n^2 * k), where n is the number of elements in nums and k is the number of subarrays.

Implementation

Here is a possible implementation in C++:

#include <vector>
#include <algorithm>
#include <climits>

using namespace std;

int minSumOfWidths(vector<int>& nums, int k) {
    int n = nums.size();
    // Sort the array initially
    sort(nums.begin(), nums.end());
    
    // Initialize DP table
    vector<vector<int>> dp(n + 1, vector<int>(k + 1, INT_MAX));
    
    // Base case: 0 elements with 0 subarrays has a width of 0
    dp[0][0] = 0;
    
    // Precompute widths for subarrays
    vector<vector<int>> width(n, vector<int>(n, 0));
    for (int i = 0; i < n; ++i) {
        for (int j = i; j < n; ++j) {
            width[i][j] = nums[j] - nums[i];
        }
    }
    
    // Fill DP table
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= k; ++j) {
            for (int x = 0; x < i; ++x) {
                if (dp[x][j - 1] != INT_MAX) {
                    dp[i][j] = min(dp[i][j], dp[x][j - 1] + width[x][i - 1]);
                }
            }
        }
    }
    
    // The answer is the minimum width sum for dividing n elements into k parts
    return dp[n][k];
}

This approach uses dynamic programming to store the minimum sum of widths needed to split the array into k subarrays. The width array is precomputed to ensure quick access while filling the DP table. The complexity is mainly bounded by O(n^2 * k) due to the triple nested loops.

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