Maximum Width Ramp

Maximum Width Ramp - Problem

A ramp in an integer array nums is a pair (i, j) for which i < j and nums[i] <= nums[j].

The width of such a ramp is j - i.

Given an integer array nums, return the maximum width of a ramp in nums.

If there is no ramp in nums, return 0.

Input & Output

Example 1 — Basic Ramp

$ Input: nums = [6,0,8,2,1,5]

› Output: 4

💡 Note: The maximum width ramp is from index 1 to index 5: nums[1] = 0 <= nums[5] = 5, width = 5 - 1 = 4

Example 2 — Decreasing Array

$ Input: nums = [9,8,1,0,1,9,4,0,4,1]

› Output: 7

💡 Note: The maximum width ramp is from index 2 to index 9: nums[2] = 1 <= nums[9] = 1, width = 9 - 2 = 7

Example 3 — No Valid Ramp

$ Input: nums = [10,9,8,7,6]

› Output: 0

💡 Note: Array is strictly decreasing, so no valid ramp exists where nums[i] <= nums[j] with i < j

Constraints

2 ≤ nums.length ≤ 5 × 10⁴
0 ≤ nums[i] ≤ 5 × 10⁴

Visualization

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Asked in

G Google 15 f Facebook 12 a Amazon 8

The key insight is to use a monotonic decreasing stack to efficiently identify potential starting positions for ramps. The optimal approach uses a monotonic stack to achieve O(n) time complexity. Time: O(n), Space: O(n)

Common Approaches

✓ Brute Force - Check All Pairs

⏱️ Time: O(n²) Space: O(1)

For each index i, check all indices j > i to see if nums[i] <= nums[j]. Keep track of the maximum width j - i found.

Monotonic Decreasing Stack

⏱️ Time: O(n) Space: O(n)

Build a monotonic decreasing stack of indices with decreasing values. Then traverse from right to left to find the maximum width ramp by popping from stack when a valid ramp is found.

Brute Force - Check All Pairs — Algorithm Steps

Iterate through each index i from 0 to n-2
For each i, check all indices j from i+1 to n-1
If nums[i] <= nums[j], update maxWidth with max(maxWidth, j-i)

Visualization

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Step-by-Step Walkthrough

Check Pairs

Compare every i < j where nums[i] <= nums[j]

Track Width

Calculate j - i for valid pairs

Find Maximum

Keep track of the maximum width found

Code -

solution.c — C

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

int solution(int* nums, int n) {
    int maxWidth = 0;
    
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            if (nums[i] <= nums[j]) {
                int width = j - i;
                if (width > maxWidth) {
                    maxWidth = width;
                }
            }
        }
    }
    
    return maxWidth;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops check all pairs of indices

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Check All Pairs — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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