Maximum Value of an Ordered Triplet II

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: 77

💡 Note: The best triplet is (0,2,4): (nums[0] - nums[2]) * nums[4] = (12 - 1) * 7 = 77.

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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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

✓ Brute Force - Triple Nested Loop

⏱️ Time: O(n³) Space: O(1)

For every valid combination of three indices, calculate (nums[i] - nums[j]) * nums[k] and track the maximum value. Return 0 if all values are negative.

Greedy - Single Pass Optimization

⏱️ Time: O(n) Space: O(1)

In one pass, maintain the maximum value of (nums[i] - nums[j]) seen so far for i < j, then for each position k, multiply this max difference by nums[k] to get potential triplet values.

Optimized - Two Pass with Prefix/Suffix Max

⏱️ Time: O(n) Space: O(n)

For each middle position j, find the maximum nums[i] before j and maximum nums[k] after j. Then calculate max difference multiplied by max suffix for optimal triplets.

Brute Force - Triple Nested Loop — Algorithm Steps

Iterate through all i from 0 to n-3
For each i, iterate j from i+1 to n-2
For each j, iterate k from j+1 to n-1
Calculate (nums[i] - nums[j]) * nums[k]
Track maximum value

Visualization

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Step-by-Step Walkthrough

Triple Loop

Generate all combinations where i < j < k

Calculate

For each triplet, compute (nums[i] - nums[j]) * nums[k]

Track Max

Keep the maximum value found

Code -

solution.c — C

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

long long solution(int* nums, int numsSize) {
    long long maxValue = LLONG_MIN;
    
    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++) {
                long long value = (long long)(nums[i] - nums[j]) * nums[k];
                if (value > maxValue) {
                    maxValue = value;
                }
            }
        }
    }
    
    return maxValue < 0 ? 0 : maxValue;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

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(n³)

Triple nested loops iterate through all combinations

✓ Linear Growth

Space Complexity

O(1)

Only using variables to track maximum

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Triple Nested Loop — Algorithm Steps

Visualization

Code -

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