Leetcode 2908: Minimum Sum of Mountain Triplets I
You are given an array of integers called nums. A mountain triplet consists of three indices (i, j, k) such that i < j < k, nums[i] < nums[j], and nums[k] < nums[j]. Your task is to return the minimum possible sum of any such mountain triplet. If no such triplet exists, return -1.
Problem
Approach
Steps
Complexity
Input: You are given an integer array nums of length n.
Example: nums = [7, 4, 8, 5, 2]
Constraints:
• 3 <= nums.length <= 50
• 1 <= nums[i] <= 50
Output: Return the minimum sum of a valid mountain triplet (i, j, k), or -1 if no such triplet exists.
Example: For the input [7, 4, 8, 5, 2], the output should be 14.
Constraints:
Goal: The goal is to identify the minimum sum of a valid mountain triplet in the array.
Steps:
• Iterate through the array with three pointers: one for the left part (i), one for the peak (j), and one for the right part (k).
• Check if the triplet formed by the indices (i, j, k) satisfies the mountain condition: nums[i] < nums[j] and nums[k] < nums[j].
• If the condition holds, calculate the sum of nums[i] + nums[j] + nums[k], and track the minimum sum found.
• Return the minimum sum if any valid triplet is found, or -1 if no such triplet exists.
Goal: The array size and values are within manageable limits for a brute force solution.
Steps:
• 3 <= nums.length <= 50
• 1 <= nums[i] <= 50
Assumptions:
• The input will always have at least three numbers.
• The solution should work for all values within the given constraints.
• Input: nums = [7, 4, 8, 5, 2]
• Explanation: In this case, the mountain triplet (1, 2, 3) is valid because 4 < 8 and 5 < 8. The sum of this triplet is 7 + 8 + 5 = 14, which is the minimum sum.
Approach: The approach involves brute force checking of all triplets (i, j, k) to find the smallest valid sum.
Observations:
• We need to check every combination of triplets that satisfy the mountain condition.
• A brute-force solution will work within the problem's constraints since the array size is relatively small.
Steps:
• Initialize a variable to store the minimum sum as INT_MAX.
• Iterate through the array using three nested loops to generate all possible triplets (i, j, k).
• For each triplet, check if it satisfies the condition: nums[i] < nums[j] and nums[k] < nums[j].
• If the condition holds, calculate the sum and update the minimum sum if the current sum is smaller.
• Return the minimum sum if a valid triplet is found, otherwise return -1.
Empty Inputs:
• The input array will always have at least three elements.
Large Inputs:
• The solution should work efficiently within the given constraints, which are n <= 50.
Special Values:
• Consider the case where no valid triplet exists in the array.
Constraints:
• All values of nums will be between 1 and 50.
int minimumSum(vector<int>& nums) {
int n=nums.size();
int ans=INT_MAX;
for(int i=0;i<n-2;i++){
for(int j=i+1;j<n-1;j++){
for(int k=j+1;k<n;k++){
if(nums[i] < nums[j] && nums[k] < nums[j]) ans = min(ans,nums[i]+nums[j]+nums[k]);
}
}
}
return (ans==INT_MAX ? -1:ans);
}
1 : Function Definition
int minimumSum(vector<int>& nums) {
Defines the function `minimumSum` which takes a vector of integers `nums` and returns the minimum sum of three distinct elements satisfying the given conditions.
2 : Variable Initialization
int n=nums.size();
Initializes `n` with the size of the input vector `nums`.
3 : Variable Initialization
int ans=INT_MAX;
Initializes `ans` to `INT_MAX` to represent the initial state where no valid triplet has been found.
4 : Triple Nested Loop
for(int i=0;i<n-2;i++){
Starts the outer loop, iterating through each element of the vector except the last two.
5 : Triple Nested Loop
for(int j=i+1;j<n-1;j++){
Starts the second loop, iterating through the elements of the vector starting from the element after `i`.
6 : Triple Nested Loop
for(int k=j+1;k<n;k++){
Starts the third loop, iterating through the elements of the vector starting from the element after `j`.
7 : Conditional Check
if(nums[i] < nums[j] && nums[k] < nums[j]) ans = min(ans,nums[i]+nums[j]+nums[k]);
Checks if `nums[i]` is smaller than `nums[j]` and `nums[k]` is smaller than `nums[j]`, and if true, updates `ans` with the minimum sum of the triplet.
8 : Return Statement
return (ans==INT_MAX ? -1:ans);
Returns the value of `ans`, which is either the minimum sum found or `-1` if no valid triplet exists.
Best Case: O(n^3)
Average Case: O(n^3)
Worst Case: O(n^3)
Description: The algorithm checks every combination of triplets, which results in a cubic time complexity.
Best Case: O(1)
Worst Case: O(1)
Description: No additional space is required apart from a few variables to track the minimum sum.
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