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Leetcode 2908. Minimum Sum of Mountain Triplets I Sure! Let’s break this problem down step by step.

Problem Statement

Given an array nums of positive integers, find the minimum sum of elements from three subarrays L, M, and N of length X, Y, and Z respectively, such that L comes before M, and M comes before N. More formally, the sum of elements in the subarrays should be minimized and the subarrays should be non-overlapping with the order constraint L < M < N.

Clarifying Questions

  1. Input Constraints:
    • What is the maximum length of the input array nums?
    • What are typical values for X, Y, and Z?
  2. Edge Cases:
    • What should we return if no valid triplet exists (for example, if the array is too short to accommodate L, M, and N)?
  3. Output:
    • Should the function return the sum of the triplets or the minimum possible sum?

Assuming the constraints are typical for a leetcode problem, let’s go ahead with the implementation:

Strategy

  1. Prefix Sums: Calculate the prefix sums of the array to make the calculation of any subarray sum more efficient.
  2. Sliding Window: Use a sliding window approach to find possible sums of subarrays of lengths X, Y, and Z.

Steps

  1. Generate Prefix Sums: This allows us to calculate the sum of any subarray in constant time.
  2. Find Minimum Sums:
    • Iterate through possible starting positions of the subarray N and for each, find the optimal positions of M and L.
    • Maintain running minimum values for possible subarray sums ending before the current M and M.

Code

Here’s how it can be implemented in C++:

#include <vector>
#include <numeric>
#include <climits>
#include <iostream>

int minSumMountainTriplets(std::vector<int>& nums, int X, int Y, int Z) {
    int n = nums.size();
    if (n < X + Y + Z) return -1;  // Not enough elements for the triplets
    
    // Prefix sums
    std::vector<int> prefix(n + 1, 0);
    for (int i = 0; i < n; ++i) {
        prefix[i + 1] = prefix[i] + nums[i];
    }
    
    auto subarraySum = [&](int l, int r) {
        return prefix[r + 1] - prefix[l];
    };

    // Define minimum sums for subarrays ending before i
    std::vector<int> minLToI(n, INT_MAX);
    std::vector<int> minLPlusIToI(n, INT_MAX);

    // Fill `minLToI` array for all valid positions of L
    for (int i = X - 1; i < n; ++i) {
        int sumL = subarraySum(i - X + 1, i);
        if (i == X - 1) {
            minLToI[i] = sumL;
        } else {
            minLToI[i] = std::min(minLToI[i - 1], sumL);
        }
    }

    // Fill `minLPlusIToI` array for all valid positions of M (i is end of M)
    for (int i = X + Y - 1; i < n; ++i) {
        int sumM = subarraySum(i - Y + 1, i);
        int minLPlusSumM = minLToI[i - Y] + sumM;
        if (i == X + Y - 1) {
            minLPlusIToI[i] = minLPlusSumM;
        } else {
            minLPlusIToI[i] = std::min(minLPlusIToI[i - 1], minLPlusSumM);
        }
    }

    // Find the optimal sum by checking positions of N
    int result = INT_MAX;
    for (int i = X + Y + Z - 1; i < n; ++i) {
        int sumN = subarraySum(i - Z + 1, i);
        result = std::min(result, minLPlusIToI[i - Z] + sumN);
    }

    return result == INT_MAX ? -1 : result;
}

int main() {
    std::vector<int> nums = {1, 2, 3, 4, 5, 6, 7, 8, 9};
    int X = 2, Y = 3, Z = 2;
    std::cout << "Minimum Sum of Mountain Triplets: " << minSumMountainTriplets(nums, X, Y, Z) << std::endl;
    return 0;
}

Time Complexity

So the overall time complexity is O(n).

This approach ensures we efficiently find the minimum sum of mountain triplets according to the given constraints.

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