Minimum Sum of Mountain Triplets I - Practice Coding Problems

Minimum Sum of Mountain Triplets I - Problem

Array Easy

You are given a 0-indexed array nums of integers.

A triplet of indices (i, j, k) is a mountain if:

i < j < k
nums[i] < nums[j] and nums[k] < nums[j]

Return the minimum possible sum of a mountain triplet of nums. If no such triplet exists, return -1.

Input & Output

Example 1 — Basic Mountain

$ Input: nums = [8,6,1,5,3]

› Output: 9

💡 Note: Triplet (2,3,4): nums[2]=1 < nums[3]=5 and nums[4]=3 < nums[3]=5. Sum = 1+5+3 = 9

Example 2 — Multiple Mountains

$ Input: nums = [5,4,8,7,10,2]

› Output: 13

💡 Note: Triplet (1,2,3): nums[1]=4 < nums[2]=8 and nums[3]=7 < nums[2]=8. Sum = 4+8+7 = 19. But (1,4,5): 4 < 10 and 2 < 10 gives 4+10+2 = 16. Best is (0,4,5): 5+10+2 = 17. Actually (0,2,5): 5+8+2 = 15. Wait, let me recalculate... (1,4,5): 4+10+2 = 16, but we need i=1,j=4,k=5, so 4<10 and 2<10 ✓. Actually the minimum is (0,2,5): 5+8+2 = 15, but 2 is not < 8. Let me check (1,2,5): 4+8+2 = 14, but 2 not < 8. The valid one is (1,4,5): 4+10+2 = 16. But actually (0,2,3): 5+8+7 = 20, but 7 not < 8. Let me try (0,4,5): 5+10+2 = 17, but 5 not < 10. Actually (1,4,5): nums[1]=4 < nums[4]=10 ✓ and nums[5]=2 < nums[4]=10 ✓, so 4+10+2=16. But let me find smaller... (1,2,5): nums[1]=4 < nums[2]=8 ✓ and nums[5]=2 < nums[2]=8 ✓, so 4+8+2=14. But wait, is 2<8? Yes. But position 5 has value 2, position 2 has value 8. So sum is 4+8+2=14. But that doesn't match expected 13. Let me recalculate the array indexing...

Example 3 — No Mountain

$ Input: nums = [6,5,4,3,4,5]

› Output: -1

💡 Note: No valid mountain triplet exists where middle element is strictly greater than both left and right elements

Constraints

3 ≤ nums.length ≤ 50
1 ≤ nums[i] ≤ 50

Visualization

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Understanding the Visualization

Input Array

Find indices i < j < k where nums[i] < nums[j] > nums[k]

Mountain Pattern

Middle element must be greater than both sides

Minimum Sum

Return smallest sum among all valid mountain triplets

Key Takeaway

🎯 Key Insight: For each potential peak position, find the minimum valid left and right elements to minimize the mountain sum

Asked in

a Amazon 15 M Microsoft 8

The key insight is to recognize that for each potential middle element (peak), we need the smallest possible left and right elements. The brute force O(n³) approach checks all triplets, while the optimized O(n²) approach precomputes minimum left values. Best approach uses two passes: Time: O(n²), Space: O(n).

Common Approaches

Approach	Time	Space	Notes
✓ Two-Pass Optimization	O(n²)	O(n)	For each middle element, precompute minimum left and find minimum right
Triple Nested Loop	O(n³)	O(1)	Check all possible triplets (i,j,k) where i < j < k

Two-Pass Optimization — Algorithm Steps

Precompute minimum left element for each position
For each middle position j, scan right for minimum valid k
Calculate sum only for valid mountain triplets
Track overall minimum sum

Visualization

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Step-by-Step Walkthrough

Preprocess

Find minimum left element for each position

Scan Right

For each middle, find minimum right element

Combine

Calculate sum for valid mountain triplets

Code -

solution.c — C

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

int solution(int* nums, int numsSize) {
    if (numsSize < 3) return -1;
    
    // Precompute minimum left values
    int* minLeft = (int*)malloc(numsSize * sizeof(int));
    for (int i = 0; i < numsSize; i++) {
        minLeft[i] = INT_MAX;
    }
    
    for (int j = 1; j < numsSize; j++) {
        for (int i = 0; i < j; i++) {
            if (nums[i] < nums[j]) {
                if (nums[i] < minLeft[j]) {
                    minLeft[j] = nums[i];
                }
            }
        }
    }
    
    int minSum = INT_MAX;
    
    // For each middle position
    for (int j = 1; j < numsSize - 1; j++) {
        if (minLeft[j] == INT_MAX) continue;
        
        // Find minimum right element
        int minRight = INT_MAX;
        for (int k = j + 1; k < numsSize; k++) {
            if (nums[k] < nums[j]) {
                if (nums[k] < minRight) {
                    minRight = nums[k];
                }
            }
        }
        
        if (minRight != INT_MAX) {
            int sum = minLeft[j] + nums[j] + minRight;
            if (sum < minSum) {
                minSum = sum;
            }
        }
    }
    
    free(minLeft);
    return minSum == INT_MAX ? -1 : minSum;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int size = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != '\n') {
            nums[size++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    int result = solution(nums, size);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Outer loop for j positions, inner loop scans right positions for each j

⚠ Quadratic Growth

Space Complexity

O(n)

Array to store minimum left values for each position

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Two-Pass Optimization — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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