Codestin Search App

Kth Largest Element in an Array - Problem

Imagine you're organizing a sports tournament and need to quickly find the k-th best performer from a list of scores. This problem asks you to find the k-th largest element in an unsorted integer array.

Given an integer array nums and an integer k, return the k-th largest element in the array. Note that we want the k-th largest element in sorted order, not the k-th distinct element.

Challenge: Can you solve this without fully sorting the array? There are elegant solutions that can do better than O(n log n) time complexity!

Example: In array [3,2,1,5,6,4] with k=2, the 2nd largest element is 5 (since sorted array would be [6,5,4,3,2,1]).

Input & Output

example_1.py — Basic Case

$ Input: nums = [3,2,1,5,6,4], k = 2

› Output: 5

💡 Note: The 2nd largest element in the sorted array [6,5,4,3,2,1] is 5

example_2.py — With Duplicates

$ Input: nums = [3,2,3,1,2,4,5,5,6], k = 4

› Output: 4

💡 Note: Sorted array: [6,5,5,4,3,3,2,2,1]. The 4th largest element is 4

example_3.py — Single Element

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

› Output: 1

💡 Note: Only one element exists, so the 1st (and only) largest element is 1

Constraints

1 ≤ k ≤ nums.length ≤ 10⁵
-10⁴ ≤ nums[i] ≤ 10⁴
k is guaranteed to be valid (within array bounds)

Visualization

Tap to expand

Understanding the Visualization

Choose Strategy

Select between full sorting O(n log n), min-heap O(n log k), or quickselect O(n) average

Partition/Process

Apply chosen strategy - sort all, maintain k-size heap, or partition around pivots

Extract Result

Return the k-th largest element from sorted array, heap root, or pivot position

Key Takeaway

🎯 Key Insight: You don't need to sort everything to find one element. Quickselect partitions strategically around pivots, eliminating half the search space each time, achieving O(n) average performance!

Asked in

f Facebook 85 a Amazon 72 Apple 68 G Google 61 ⊞ Microsoft 55 in LinkedIn 43

The optimal approach uses Quickselect algorithm with average O(n) time complexity. By partitioning around pivot elements and recursing only on the side containing the k-th largest element, we avoid unnecessary sorting. Alternative approaches include Min-Heap (O(n log k)) which is efficient when k is small, and full sorting (O(n log n)) for simplicity.

Common Approaches

Approach	Time	Space	Notes
✓ Quickselect Algorithm (Optimal)	O(n) average, O(n²) worst case	O(1)	Use divide-and-conquer partitioning to find k-th largest in average O(n) time
Min-Heap Approach	O(n log k)	O(k)	Use a min-heap of size k to track the k largest elements
Brute Force (Full Sorting)	O(n log n)	O(1)	Sort the entire array and return the k-th element from the end

Quickselect Algorithm (Optimal) — Algorithm Steps

Choose a pivot element (can be random or median-of-medians)
Partition array around pivot: elements larger than pivot on left, smaller on right
If pivot is at position k-1, return pivot (found k-th largest)
If pivot position > k-1, recurse on left subarray
If pivot position < k-1, recurse on right subarray with adjusted k

Visualization

Tap to expand

Step-by-Step Walkthrough

Choose Pivot

Select pivot (e.g., 5) and partition array

Partition

Move larger elements left, smaller elements right

Recurse

Focus only on the side containing k-th largest

Code -

solution.c — C

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

void swap(int* a, int* b) {
    int temp = *a;
    *a = *b;
    *b = temp;
}

int partition(int* nums, int left, int right, int pivotIdx) {
    int pivotValue = nums[pivotIdx];
    swap(&nums[pivotIdx], &nums[right]);
    
    int storeIdx = left;
    for (int i = left; i < right; i++) {
        if (nums[i] > pivotValue) {
            swap(&nums[i], &nums[storeIdx]);
            storeIdx++;
        }
    }
    
    swap(&nums[storeIdx], &nums[right]);
    return storeIdx;
}

int quickselect(int* nums, int left, int right, int kSmallest) {
    if (left == right) return nums[left];
    
    int pivotIdx = left + rand() % (right - left + 1);
    pivotIdx = partition(nums, left, right, pivotIdx);
    
    if (kSmallest == pivotIdx) {
        return nums[kSmallest];
    } else if (kSmallest < pivotIdx) {
        return quickselect(nums, left, pivotIdx - 1, kSmallest);
    } else {
        return quickselect(nums, pivotIdx + 1, right, kSmallest);
    }
}

int findKthLargest(int* nums, int numsSize, int k) {
    srand(time(NULL));
    return quickselect(nums, 0, numsSize - 1, k - 1);
}

int main() {
    int nums[] = {3,2,1,5,6,4};
    int k = 2;
    printf("%d\n", findKthLargest(nums, 6, k));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n) average, O(n²) worst case

Average case: each partition reduces problem size by half. Worst case: poor pivot choices lead to O(n²)

⚠ Quadratic Growth

Space Complexity

O(1)

In-place algorithm, only uses constant extra space for recursion variables

✓ Linear Space

89.5K Views

Very High Frequency

~15 min Avg. Time

2.8K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Quickselect Algorithm (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler