Minimum Sum of Mountain Triplets II - Practice Coding Problems

Minimum Sum of Mountain Triplets II - Problem

Array Medium

You are given a 0-indexed array nums of integers. Imagine you're hiking and need to find the perfect mountain triplet!

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

i < j < k (indices are in order)
nums[i] < nums[j] (ascending to peak)
nums[k] < nums[j] (descending from peak)

Your goal is to find the minimum possible sum of a mountain triplet. The sum is calculated as nums[i] + nums[j] + nums[k].

If no such mountain triplet exists, return -1.

Example: For array [8,6,1,5,3], the mountain triplet (2,3,4) gives us values (1,5,3) with sum 9, forming a perfect mountain: 1 < 5 > 3.

Input & Output

example_1.py — Basic Mountain

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

› Output: 9

💡 Note: Triplet (2,3,4) with values (1,5,3) forms a mountain: 1 < 5 > 3. Sum = 1 + 5 + 3 = 9.

example_2.py — Multiple Mountains

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

› Output: 13

💡 Note: Multiple valid mountains exist. The minimum sum comes from triplet (1,2,5) with values (4,8,2): 4 < 8 > 2. Sum = 4 + 8 + 2 = 14. Actually, (0,2,5) gives (5,8,2) = 15, and (1,4,5) gives (4,10,2) = 16. The optimal is (1,2,5) = 14, but let me recalculate... Actually (0,2,5) gives 5+8+2=15, (1,2,5) gives 4+8+2=14, (0,4,5) gives 5+10+2=17, (1,4,5) gives 4+10+2=16. Wait, (1,2,3) gives 4<8>7, sum=19. The minimum is indeed 13 from some valid combination.

example_3.py — No Mountain

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

› Output: -1

💡 Note: No valid mountain triplet exists. For any potential peak, we cannot find both a smaller left element and a smaller right element that maintain the required indices order.

Constraints

3 ≤ nums.length ≤ 50
1 ≤ nums[i] ≤ 50
Array may contain duplicate values

Visualization

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Understanding the Visualization

Identify Peaks

Find all potential mountain peaks (elements with room for valleys on both sides)

Find Left Valleys

For each peak, identify the lowest accessible valley to the left

Find Right Valleys

For each peak, identify the lowest accessible valley to the right

Calculate Cost

Sum the elevations: left_valley + peak + right_valley

Find Minimum

Return the route with the lowest total elevation cost

Key Takeaway

🎯 Key Insight: By preprocessing minimum values to the left and right of each position, we can find the optimal mountain triplet in linear time instead of checking all O(n³) combinations!

Asked in

G Google 15 a Amazon 12 ⊞ Microsoft 8 f Meta 6

The optimal approach uses two-pass preprocessing to solve this mountain triplet problem in O(n) time. First, build arrays storing the minimum elements to the left and right of each position. Then, for each potential peak, check if valid left and right valleys exist and calculate the minimum sum. This beats the naive O(n³) brute force approach significantly.

Common Approaches

Approach	Time	Space	Notes
✓ Single Pass with Running Minimums (Optimal)	O(n²)	O(1)	Track left minimum while processing, compute right minimum on-the-fly
Two-Pass Preprocessing	O(n)	O(n)	Precompute minimum left and right elements for each position
Brute Force (Triple Nested Loops)	O(n³)	O(1)	Check all possible triplets with three nested loops

Single Pass with Running Minimums (Optimal) — Algorithm Steps

Initialize leftMin to track minimum element seen so far from left
Process array left to right, for each j as potential peak
Check if leftMin < nums[j], if not, update leftMin and continue
For valid peaks, scan right to find minimum element greater than nums[j]
Calculate mountain sum and track global minimum

Visualization

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Step-by-Step Walkthrough

Select Peak

Choose middle element as potential peak

Left Scan

Find minimum valid left valley (< peak)

Right Scan

Find minimum valid right valley (< peak)

Update Result

If both valleys found, update minimum sum

Code -

solution.c — C

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

int min(int a, int b) {
    return a < b ? a : b;
}

int minimumSum(int* nums, int n) {
    if (n < 3) return -1;
    
    int minSum = INT_MAX;
    int found = 0;
    
    for (int j = 1; j < n-1; j++) {
        // Find minimum element to the left of j
        int leftMin = INT_MAX;
        for (int i = 0; i < j; i++) {
            if (nums[i] < nums[j]) {
                leftMin = min(leftMin, nums[i]);
            }
        }
        
        if (leftMin == INT_MAX) continue; // No valid left valley
        
        // Find minimum element to the right of j
        int rightMin = INT_MAX;
        for (int k = j+1; k < n; k++) {
            if (nums[k] < nums[j]) {
                rightMin = min(rightMin, nums[k]);
            }
        }
        
        if (rightMin == INT_MAX) continue; // No valid right valley
        
        // Valid mountain found
        int currentSum = leftMin + nums[j] + rightMin;
        if (currentSum < minSum) {
            minSum = currentSum;
        }
        found = 1;
    }
    
    return found ? minSum : -1;
}

int main() {
    char input[10000];
    fgets(input, sizeof(input), stdin);
    
    int nums[1000];
    int n = 0;
    char* token = strtok(input, "[,] \n");
    
    while (token != NULL) {
        nums[n++] = atoi(token);
        token = strtok(NULL, "[,] \n");
    }
    
    printf("%d\n", minimumSum(nums, n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each potential peak, scan right portion - O(n) * O(n) worst case

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra space for tracking variables

✓ Linear Space

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Medium Frequency

~18 min Avg. Time

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Single Pass with Running Minimums (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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