Find the Most Competitive Subsequence - Practice Coding Problems

Find the Most Competitive Subsequence - Problem

Given an integer array nums and a positive integer k, return the most competitive subsequence of nums of size k.

An array's subsequence is a resulting sequence obtained by erasing some (possibly zero) elements from the array.

We define that a subsequence a is more competitive than a subsequence b (of the same length) if in the first position where a and b differ, subsequence a has a number less than the corresponding number in b. For example, [1,3,4] is more competitive than [1,3,5] because the first position they differ is at the final number, and 4 < 5.

Input & Output

Example 1 — Basic Case

$ Input: nums = [3,5,2,6], k = 2

› Output: [2,6]

💡 Note: Among all possible subsequences of length 2: [3,5], [3,2], [3,6], [5,2], [5,6], [2,6]. The subsequence [2,6] is most competitive because 2 < 3 (first elements compared).

Example 2 — Larger Array

$ Input: nums = [2,4,3,3,5,4,9,6], k = 4

› Output: [2,3,3,4]

💡 Note: We need to select 4 elements maintaining order. The most competitive subsequence starts with the smallest possible elements: 2, then 3, then 3, then 4.

Example 3 — All Elements

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

› Output: [1,2,3]

💡 Note: When k equals array length, we must take all elements in their original order.

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ k ≤ nums.length
1 ≤ nums[i] ≤ 10⁹

Visualization

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Understanding the Visualization

Input

Array [3,5,2,6] with k=2 (need 2 elements)

Process

Use monotonic stack to greedily select smallest elements

Output

Most competitive subsequence [2,6]

Key Takeaway

🎯 Key Insight: Use monotonic stack to greedily build the lexicographically smallest subsequence by removing larger elements when possible

Asked in

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

The key insight is to use a monotonic stack that maintains the smallest possible elements while ensuring we can still select k elements total. The optimal approach uses a greedy strategy with O(n) time complexity by removing larger elements when beneficial.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Generate All Subsequences	O(C(n,k) × k)	O(C(n,k) × k)	Generate all possible subsequences of size k and find the lexicographically smallest one
Greedy with Monotonic Stack	O(n)	O(k)	Use a stack to build the result greedily, removing larger elements when beneficial

Brute Force - Generate All Subsequences — Algorithm Steps

Generate all possible subsequences of size k
Compare each subsequence lexicographically
Return the most competitive one

Visualization

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Step-by-Step Walkthrough

Generate

Create all possible subsequences of length k

Compare

Compare each subsequence lexicographically

Select

Return the most competitive one

Code -

solution.c — C

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

int* best;
int bestSize = 0;
int first = 1;

int isLexicographicallySmaller(int* a, int* b, int size) {
    for (int i = 0; i < size; i++) {
        if (a[i] < b[i]) return 1;
        if (a[i] > b[i]) return 0;
    }
    return 0;
}

void generateCombinations(int* nums, int n, int k, int start, int* current, int currentSize) {
    if (currentSize == k) {
        int* subsequence = malloc(k * sizeof(int));
        for (int i = 0; i < k; i++) {
            subsequence[i] = nums[current[i]];
        }
        
        if (first || isLexicographicallySmaller(subsequence, best, k)) {
            if (best) free(best);
            best = subsequence;
            bestSize = k;
            first = 0;
        } else {
            free(subsequence);
        }
        return;
    }
    
    for (int i = start; i <= n - (k - currentSize); i++) {
        current[currentSize] = i;
        generateCombinations(nums, n, k, i + 1, current, currentSize + 1);
    }
}

int* solution(int* nums, int n, int k) {
    best = NULL;
    bestSize = 0;
    first = 1;
    
    int* current = malloc(k * sizeof(int));
    generateCombinations(nums, n, k, 0, current, 0);
    free(current);
    
    return best;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[1000];
    int n = 0;
    char* token = strtok(line + 1, ",]");
    while (token) {
        nums[n++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int k;
    scanf("%d", &k);
    
    int* result = solution(nums, n, k);
    
    printf("[");
    for (int i = 0; i < k; i++) {
        if (i > 0) printf(",");
        printf("%d", result[i]);
    }
    printf("]\n");
    
    if (result) free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(C(n,k) × k)

Generate C(n,k) combinations and compare each of length k

✓ Linear Growth

Space Complexity

O(C(n,k) × k)

Store all possible subsequences of length k

⚡ Linearithmic Space

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Medium Frequency

~25 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Generate All Subsequences — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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