LeetCode #215 — MEDIUM

Kth Largest Element in an Array

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

Given an integer array nums and an integer k, return the k^th largest element in the array.

Note that it is the k^th largest element in the sorted order, not the k^th distinct element.

Can you solve it without sorting?

Example 1:

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

Example 2:

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

Constraints:

1 <= k <= nums.length <= 10⁵
-10⁴ <= nums[i] <= 10⁴

Step 01

Brute Force Baseline

Problem summary: Given an integer array nums and an integer k, return the kth largest element in the array. Note that it is the kth largest element in the sorted order, not the kth distinct element. Can you solve it without sorting?

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[3,2,1,5,6,4]
2

Example 2

[3,2,3,1,2,4,5,5,6]
4

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #215: Kth Largest Element in an Array
class Solution {
    private int[] nums;
    private int k;

    public int findKthLargest(int[] nums, int k) {
        this.nums = nums;
        this.k = nums.length - k;
        return quickSort(0, nums.length - 1);
    }

    private int quickSort(int l, int r) {
        if (l == r) {
            return nums[l];
        }
        int i = l - 1, j = r + 1;
        int x = nums[(l + r) >>> 1];
        while (i < j) {
            while (nums[++i] < x) {
            }
            while (nums[--j] > x) {
            }
            if (i < j) {
                int t = nums[i];
                nums[i] = nums[j];
                nums[j] = t;
            }
        }
        if (j < k) {
            return quickSort(j + 1, r);
        }
        return quickSort(l, j);
    }
}

// Accepted solution for LeetCode #215: Kth Largest Element in an Array
func findKthLargest(nums []int, k int) int {
	k = len(nums) - k
	var quickSort func(l, r int) int
	quickSort = func(l, r int) int {
		if l == r {
			return nums[l]
		}
		i, j := l-1, r+1
		x := nums[(l+r)>>1]
		for i < j {
			for {
				i++
				if nums[i] >= x {
					break
				}
			}
			for {
				j--
				if nums[j] <= x {
					break
				}
			}
			if i < j {
				nums[i], nums[j] = nums[j], nums[i]
			}
		}
		if j < k {
			return quickSort(j+1, r)
		}
		return quickSort(l, j)
	}
	return quickSort(0, len(nums)-1)
}

# Accepted solution for LeetCode #215: Kth Largest Element in an Array
class Solution:
    def findKthLargest(self, nums: List[int], k: int) -> int:
        def quick_sort(l: int, r: int) -> int:
            if l == r:
                return nums[l]
            i, j = l - 1, r + 1
            x = nums[(l + r) >> 1]
            while i < j:
                while 1:
                    i += 1
                    if nums[i] >= x:
                        break
                while 1:
                    j -= 1
                    if nums[j] <= x:
                        break
                if i < j:
                    nums[i], nums[j] = nums[j], nums[i]
            if j < k:
                return quick_sort(j + 1, r)
            return quick_sort(l, j)

        n = len(nums)
        k = n - k
        return quick_sort(0, n - 1)

// Accepted solution for LeetCode #215: Kth Largest Element in an Array
impl Solution {
    pub fn find_kth_largest(mut nums: Vec<i32>, k: i32) -> i32 {
        let len = nums.len();
        let k = len - k as usize;
        Self::quick_sort(&mut nums, 0, len - 1, k)
    }

    fn quick_sort(nums: &mut Vec<i32>, l: usize, r: usize, k: usize) -> i32 {
        if l == r {
            return nums[l];
        }

        let (mut i, mut j) = (l as isize - 1, r as isize + 1);
        let x = nums[(l + r) / 2];

        while i < j {
            i += 1;
            while nums[i as usize] < x {
                i += 1;
            }

            j -= 1;
            while nums[j as usize] > x {
                j -= 1;
            }

            if i < j {
                nums.swap(i as usize, j as usize);
            }
        }

        let j = j as usize;
        if j < k {
            Self::quick_sort(nums, j + 1, r, k)
        } else {
            Self::quick_sort(nums, l, j, k)
        }
    }
}

// Accepted solution for LeetCode #215: Kth Largest Element in an Array
function findKthLargest(nums: number[], k: number): number {
    const n = nums.length;
    k = n - k;
    const quickSort = (l: number, r: number): number => {
        if (l === r) {
            return nums[l];
        }
        let [i, j] = [l - 1, r + 1];
        const x = nums[(l + r) >> 1];
        while (i < j) {
            while (nums[++i] < x);
            while (nums[--j] > x);
            if (i < j) {
                [nums[i], nums[j]] = [nums[j], nums[i]];
            }
        }
        if (j < k) {
            return quickSort(j + 1, r);
        }
        return quickSort(l, j);
    };
    return quickSort(0, n - 1);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(log n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.