Next Greater Element I

Next Greater Element I - Problem

The next greater element of some element x in an array is the first greater element that is to the right of x in the same array.

You are given two distinct 0-indexed integer arrays nums1 and nums2, where nums1 is a subset of nums2.

For each 0 <= i < nums1.length, find the index j such that nums1[i] == nums2[j] and determine the next greater element of nums2[j] in nums2. If there is no next greater element, then the answer for this query is -1.

Return an array ans of length nums1.length such that ans[i] is the next greater element as described above.

Input & Output

Example 1 — Basic Case

$ Input: nums1 = [4,1,2], nums2 = [1,3,4,2]

› Output: [-1,3,-1]

💡 Note: For 4: appears at index 2 in nums2, no greater element to the right → -1. For 1: appears at index 0, next greater is 3 → 3. For 2: appears at index 3, no greater element to the right → -1.

Example 2 — All Elements Found

$ Input: nums1 = [2,4], nums2 = [1,2,3,4]

› Output: [3,-1]

💡 Note: For 2: appears at index 1 in nums2, next greater is 3 → 3. For 4: appears at index 3, no greater element to the right → -1.

Example 3 — Single Element

$ Input: nums1 = [1], nums2 = [1,2]

› Output: [2]

💡 Note: For 1: appears at index 0 in nums2, next greater is 2 → 2.

Constraints

1 ≤ nums1.length ≤ nums2.length ≤ 1000
0 ≤ nums1[i], nums2[i] ≤ 10⁴
All integers in nums1 and nums2 are unique
All the integers of nums1 also appear in nums2

Visualization

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Asked in

G Google 15 a Amazon 12 f Facebook 8 M Microsoft 6

The key insight is using a monotonic decreasing stack to efficiently find next greater elements in a single pass. The optimal approach builds a hash map of next greater elements using the stack, then performs O(1) lookups for each query. Time: O(m + n), Space: O(n).

Common Approaches

✓ Brute Force - Linear Search

⏱️ Time: O(m × n) Space: O(1)

For each element in nums1, locate it in nums2, then scan rightward from that position until we find a greater element or reach the end.

Monotonic Stack - Optimal Solution

⏱️ Time: O(m + n) Space: O(n)

Build a hash map of next greater elements using a monotonic decreasing stack, then lookup each element from nums1 in the map.

Brute Force - Linear Search — Algorithm Steps

For each element in nums1
Find its position in nums2
Scan right from that position
Return first greater element or -1

Visualization

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Step-by-Step Walkthrough

Find Element

Locate nums1[i] in nums2

Search Right

Scan rightward for first greater element

Record Result

Store the next greater element or -1

Code -

solution.c — C

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

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);
    }
}

void printArray(int* arr, int size) {
    printf("[");
    for (int i = 0; i < size; i++) {
        printf("%d", arr[i]);
        if (i < size - 1) printf(",");
    }
    printf("]\n");
}

int* solution(int* nums1, int nums1Size, int* nums2, int nums2Size) {
    int* result = malloc(nums1Size * sizeof(int));
    
    for (int i = 0; i < nums1Size; i++) {
        int num = nums1[i];
        
        // Find the position of num in nums2
        int pos = -1;
        for (int j = 0; j < nums2Size; j++) {
            if (nums2[j] == num) {
                pos = j;
                break;
            }
        }
        
        int nextGreater = -1;
        // Look for next greater element to the right
        for (int j = pos + 1; j < nums2Size; j++) {
            if (nums2[j] > num) {
                nextGreater = nums2[j];
                break;
            }
        }
        
        result[i] = nextGreater;
    }
    
    return result;
}

int main() {
    char line1[10000], line2[10000];
    fgets(line1, sizeof(line1), stdin);
    fgets(line2, sizeof(line2), stdin);
    
    int nums1[1000], nums2[1000];
    int size1, size2;
    
    parseArray(line1, nums1, &size1);
    parseArray(line2, nums2, &size2);
    
    int* result = solution(nums1, size1, nums2, size2);
    printArray(result, size1);
    
    free(result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(m × n)

For each of m elements in nums1, we potentially scan all n elements in nums2

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space for variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Linear Search — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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