Maximum Value of an Ordered Triplet II - Practice Coding Problems

Maximum Value of an Ordered Triplet II - Problem

Array Medium

You are given a 0-indexed integer array nums.

Return the maximum value over all triplets of indices (i, j, k) such that i < j < k. If all such triplets have a negative value, return 0.

The value of a triplet of indices (i, j, k) is equal to (nums[i] - nums[j]) * nums[k].

Input & Output

Example 1 — Basic Case

$ Input: nums = [12,6,1,2,7]

› Output: 42

💡 Note: The triplet (0,2,4) gives us (nums[0] - nums[2]) * nums[4] = (12 - 1) * 7 = 77. However, we need i < j < k, so we check (0,1,4): (12 - 6) * 7 = 42, which is the maximum.

Example 2 — All Negative Results

$ Input: nums = [1,10,3,4,19]

› Output: 133

💡 Note: The best triplet is (1,2,4): (nums[1] - nums[2]) * nums[4] = (10 - 3) * 19 = 133.

Example 3 — Return Zero Case

$ Input: nums = [1,2,3]

› Output: 0

💡 Note: The only triplet is (0,1,2): (1 - 2) * 3 = -3, which is negative, so we return 0.

Constraints

3 ≤ nums.length ≤ 10⁵
1 ≤ nums[i] ≤ 10⁶

Visualization

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Understanding the Visualization

Input

Array [12,6,1,2,7] with indices 0,1,2,3,4

Process

Check all valid triplets (i,j,k) where i < j < k

Output

Maximum value found: (12-6)*7 = 42

Key Takeaway

🎯 Key Insight: Track the maximum difference (nums[i] - nums[j]) while scanning, then multiply by each nums[k]

Asked in

G Google 12 a Amazon 8 M Microsoft 6

The key insight is to track the maximum difference (nums[i] - nums[j]) while scanning and multiply by each nums[k]. The optimal greedy approach achieves this in one pass. Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Optimized - Two Pass with Prefix/Suffix Max	O(n)	O(n)	Pre-compute maximum values before and after each position
Brute Force - Triple Nested Loop	O(n³)	O(1)	Check all possible triplets (i,j,k) where i < j < k
Greedy - Single Pass Optimization	O(n)	O(1)	Track maximum difference while scanning, multiply by current element

Optimized - Two Pass with Prefix/Suffix Max — Algorithm Steps

Pre-compute prefix maximum array for each position
Pre-compute suffix maximum array for each position
For each j, calculate (prefixMax[j] - nums[j]) * suffixMax[j]
Return maximum value found

Visualization

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Step-by-Step Walkthrough

Build Arrays

Create prefix max and suffix max arrays

Calculate

For each j: (prefixMax[j-1] - nums[j]) * suffixMax[j+1]

Return Max

Return the maximum value found

Code -

solution.c — C

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

long long solution(int* nums, int numsSize) {
    if (numsSize < 3) return 0;
    
    int* prefixMax = (int*)malloc(numsSize * sizeof(int));
    int* suffixMax = (int*)malloc(numsSize * sizeof(int));
    
    // Compute prefix max
    prefixMax[0] = nums[0];
    for (int i = 1; i < numsSize; i++) {
        prefixMax[i] = prefixMax[i-1] > nums[i] ? prefixMax[i-1] : nums[i];
    }
    
    // Compute suffix max
    suffixMax[numsSize-1] = nums[numsSize-1];
    for (int i = numsSize-2; i >= 0; i--) {
        suffixMax[i] = suffixMax[i+1] > nums[i] ? suffixMax[i+1] : nums[i];
    }
    
    long long maxValue = 0;
    for (int j = 1; j < numsSize-1; j++) {
        long long value = (long long)(prefixMax[j-1] - nums[j]) * suffixMax[j+1];
        if (value > maxValue) {
            maxValue = value;
        }
    }
    
    free(prefixMax);
    free(suffixMax);
    return maxValue;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    int nums[1000];
    int numsSize = 0;
    
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    long long result = solution(nums, numsSize);
    printf("%lld\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Three linear passes through the array

✓ Linear Growth

Space Complexity

O(n)

Two additional arrays for prefix and suffix maximums

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Optimized - Two Pass with Prefix/Suffix Max — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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