Minimum Sum of Mountain Triplets I - Practice Coding Problems

Minimum Sum of Mountain Triplets I - Problem

Array Easy

Imagine you're a mountain climber looking for the perfect mountain triplet - three peaks where the middle one towers above the others! 🏔️

Given a 0-indexed array nums of integers, you need to find a triplet of indices (i, j, k) that forms a "mountain" pattern where:

i < j < k (indices are in order)
nums[i] < nums[j] (left peak is shorter than middle)
nums[k] < nums[j] (right peak is shorter than middle)

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

Example: In array [8,6,1,5,3], the triplet (2,3,4) gives us values (1,5,3) - a perfect mountain with sum 9!

Input & Output

example_1.py — Basic Mountain

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

› Output: 9

💡 Note: We can form a mountain with indices (2,3,4): values (1,5,3). Since 1 < 5 and 3 < 5, this is a valid mountain with sum 1+5+3 = 9. This is the minimum possible sum.

example_2.py — Multiple Mountains

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

› Output: 13

💡 Note: Several mountain triplets exist: (1,2,3) gives (4,8,7) with sum 19, (0,4,5) gives (5,10,2) with sum 17, and (1,4,5) gives (4,10,2) with sum 16. But the minimum is (0,2,5) giving (5,8,2) with sum 15. Wait, let me recalculate: (0,2,5) gives 5+8+2=15, but we need to check if this forms a valid mountain: 5 < 8 ✓ and 2 < 8 ✓. Actually, the minimum valid mountain is (1,2,5) giving 4+8+2=14. Let me verify: 4 < 8 ✓ and 2 < 8 ✓. The answer should be 13 from triplet (0,2,3): 5+8+7=20? No, that's wrong. Let me recalculate properly: (1,4,5) = 4+10+2=16, (1,2,5) = 4+8+2=14, (0,4,5) = 5+10+2=17. The minimum valid mountain appears to be (1,2,5) = 4+8+2=14. The expected output suggests 13, so there must be another valid combination I missed.

example_3.py — No Mountain

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

› Output: -1

💡 Note: Let's check all possible mountains: For any middle element to form a mountain, we need a smaller element on both sides. Looking at the array [6,5,4,3,4,5], we can find valid mountains like (2,4,5) giving (4,4,5) - but 4 is not less than 4, so invalid. Actually, (1,4,5) gives (5,4,5) which is invalid since 5 ≮ 4. Let me check (0,4,5): (6,4,5) - invalid since 6 ≮ 4. After checking systematically, there are valid mountains like (2,5,4) but that violates i < j < k. The correct triplet (1,5,4) is invalid due to index order. Actually, (3,4,5) gives indices in order with values (3,4,5): 3 < 4 ✓ and 5 ≮ 4 ✗. It appears no valid mountain exists, hence -1.

Visualization

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Understanding the Visualization

Survey the Terrain

Fly over the mountain range [8,6,1,5,3] and identify potential middle peaks

Check Peak Validity

For each potential middle peak, verify it's taller than at least one mountain on each side

Find Minimum Costs

For valid middle peaks, find the cheapest left and right mountains to minimize total expedition cost

Optimal Formation

The triplet (1,5,3) at indices (2,3,4) gives the minimum cost of 9 for a perfect mountain formation

Key Takeaway

🎯 Key Insight: For each potential middle peak, we only need to find the cheapest (minimum) valid mountains on both sides. This transforms a complex triplet search into an efficient two-sided minimum-finding problem!

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each position we may scan up to n elements on the right, but with better practical performance

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables to track minimum values and current position

✓ Linear Space

Constraints

3 ≤ nums.length ≤ 50
1 ≤ nums[i] ≤ 10⁸
All elements in the array are positive integers

Asked in

G Google 23 a Amazon 18 ⊞ Microsoft 15 f Meta 12

The optimal approach uses a systematic scan where for each potential middle peak, we find the minimum valid left and right elements. This achieves O(n²) time complexity with O(1) space. The key insight is that we only need the minimum elements on each side that satisfy the mountain constraint, rather than checking all possible triplet combinations.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Triple Nested Loops)	O(n³)	O(1)	Check every possible triplet (i,j,k) to find the minimum mountain sum
Two-Pass Optimization	O(n²)	O(1)	Precompute minimum left and right elements for each position to optimize search
One-Pass Optimal Solution	O(n²)	O(1)	Track minimum left element while scanning, achieving O(n²) with better constants

Brute Force (Triple Nested Loops) — Algorithm Steps

Initialize minimum sum to infinity
Use three nested loops: i from 0 to n-3, j from i+1 to n-2, k from j+1 to n-1
For each triplet, check if nums[i] < nums[j] and nums[k] < nums[j]
If valid mountain, update minimum sum
Return minimum sum or -1 if no valid mountain found

Visualization

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Step-by-Step Walkthrough

Initialize

Start with three pointers i, j, k

Check Pattern

Verify if nums[i] < nums[j] > nums[k]

Update Minimum

If valid mountain, update minimum sum

Code -

solution.c — C

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

int minimumSum(int* nums, int numsSize) {
    int minSum = INT_MAX;
    
    for (int i = 0; i < numsSize - 2; i++) {
        for (int j = i + 1; j < numsSize - 1; j++) {
            for (int k = j + 1; k < numsSize; k++) {
                if (nums[i] < nums[j] && nums[k] < nums[j]) {
                    int sum = nums[i] + nums[j] + nums[k];
                    if (sum < minSum) {
                        minSum = sum;
                    }
                }
            }
        }
    }
    
    return minSum == INT_MAX ? -1 : minSum;
}

int main() {
    int nums[1000];
    int n = 0;
    char c;
    
    scanf("%c", &c); // skip '['
    while (scanf("%d", &nums[n]) == 1) {
        n++;
        scanf("%c", &c);
        if (c == ']') break;
    }
    
    printf("%d\n", minimumSum(nums, n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Three nested loops each running up to n times in worst case

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables to track minimum sum and indices

✓ Linear Space

Constraints

3 ≤ nums.length ≤ 50
1 ≤ nums[i] ≤ 10⁸
All elements in the array are positive integers

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Triple Nested Loops) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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