Find the Kth Largest Integer in the Array solution and explanation
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The problem
You are given an array of strings nums and an integer k. Each string in nums represents an integer without leading zeros.
Return the string that represents the kth largest integer in nums.
Note: Duplicate numbers should be counted distinctly. For example, if nums is ["1","2","2"], "2" is the first largest integer, "2" is the second-largest integer, and "1" is the third-largest integer.
Example 1:
Input: nums = ["3","6","7","10"], k = 4 Output: "3" Explanation: The numbers in nums sorted in non-decreasing order are ["3","6","7","10"]. The 4th largest integer in nums is "3".
Example 2:
Input: nums = ["2","21","12","1"], k = 3 Output: "2" Explanation: The numbers in nums sorted in non-decreasing order are ["1","2","12","21"]. The 3rd largest integer in nums is "2".
Example 3:
Input: nums = ["0","0"], k = 2 Output: "0" Explanation: The numbers in nums sorted in non-decreasing order are ["0","0"]. The 2nd largest integer in nums is "0".
Constraints:
1 <= k <= nums.length <= 1041 <= nums[i].length <= 100nums[i]consists of only digits.nums[i]will not have any leading zeros.
Approach overview
Problem Overview: You are given an array of integers represented as strings. The goal is to return the kth largest integer. Since values can exceed normal integer limits, comparison must be done using string length first, then lexicographic order.
Approach 1: Sort and Pick Approach (O(n log n) time, O(1)–O(n) space)
The simplest solution sorts the array of numeric strings in descending numeric order and returns the element at index k-1. Because the values are stored as strings, comparison works by checking length first (longer string means larger number). If two strings have the same length, compare them lexicographically. Most languages allow custom comparators during sorting, making this approach straightforward. This method relies heavily on sorting and works well when the array size is moderate and simplicity is preferred.
Approach 2: Min-Heap (Priority Queue) Approach (O(n log k) time, O(k) space)
This approach maintains a min-heap of size k. Iterate through the array and push each numeric string into the heap using the same comparison logic (length first, then lexicographic). If the heap size exceeds k, remove the smallest element. After processing all elements, the root of the heap is the kth largest number. This approach avoids sorting the entire array and is more efficient when k is much smaller than n. It uses a heap (priority queue) to track the top k largest values.
Approach 3: Quickselect (Average O(n) time, O(1) space)
A selection algorithm similar to quicksort can partition the array around a pivot and recursively search the side containing the kth largest element. Instead of fully sorting the array, Quickselect only processes the portion that contains the target rank. Comparisons still follow the numeric string rules (length then lexicographic). This approach uses ideas from divide and conquer. While the average time complexity is linear, the worst case can degrade to O(n²) without randomized pivots.
Recommended for interviews: The min-heap solution is usually the safest choice. It demonstrates understanding of heaps and efficiently handles large datasets with O(n log k) complexity. Sorting shows clear reasoning and is often accepted in interviews, but the heap approach highlights stronger algorithmic thinking when k is small relative to n.
Approach 1: Sort and Pick Approach
Sort and Pick Approach: The simplest approach is to sort the array of strings based on their numeric values (considering the differences in lengths and lexicographical order when required) and then pick the kth largest element from this sorted array. This approach takes advantage of built-in sort capabilities that can be customized for string comparison in terms of numeric values. After sorting the array in descending order, the kth element can be directly accessed.
def kthLargestNumber(nums, k):
nums.sort(key=lambda x: (len(x), x), reverse=True)
return nums[k - 1]
print(kthLargestNumber(["3", "6", "7", "10"], 4)) # Output: "3"import java.util.*;
public class KthLargestNumber {
public static String kthLargestNumber(String[] nums, int k) {
Arrays.sort(nums, (a, b) -> {
if (a.length() != b.length()) return b.length() - a.length();
return b.compareTo(a);
});
return nums[k - 1];
}
public static void main(String[] args) {
String[] nums = {"3", "6", "7", "10"};
int k = 4;
System.out.println(kthLargestNumber(nums, k)); // Output: "3"
}
}#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
string kthLargestNumber(vector<string>& nums, int k) {
sort(nums.begin(), nums.end(), [](const string &a, const string &b) {
if (a.size() != b.size()) return a.size() > b.size();
return a > b;
});
return nums[k - 1];
}
int main() {
vector<string> nums = {"3", "6", "7", "10"};
int k = 4;
cout << kthLargestNumber(nums, k) << endl; // Output: "3"
return 0;
}function kthLargestNumber(nums, k) {
nums.sort((a, b) => {
if (a.length !== b.length) return b.length - a.length;
return b > a ? 1 : -1;
});
return nums[k - 1];
}
console.log(kthLargestNumber(["3", "6", "7", "10"], 4)); // Output: "3";#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int compare(const void *a, const void *b) {
char *num1 = *(char **)a;
char *num2 = *(char **)b;
int len1 = strlen(num1);
int len2 = strlen(num2);
if (len1 != len2) return len2 - len1;
return strcmp(num2, num1);
}
const char* kthLargestNumber(char **nums, int numsSize, int k) {
qsort(nums, numsSize, sizeof(char *), compare);
return nums[k - 1];
}
int main() {
char *nums[] = {"3", "6", "7", "10"};
int k = 4;
printf("%s\n", kthLargestNumber(nums, 4, k)); // Output: "3"
return 0;
}using System;
class KthLargestNumber {
public static string KthLargestNumber(string[] nums, int k) {
Array.Sort(nums, (a, b) => {
if (a.Length != b.Length) return b.Length - a.Length;
return b.CompareTo(a);
});
return nums[k - 1];
}
static void Main() {
string[] nums = {"3", "6", "7", "10"};
int k = 4;
Console.WriteLine(KthLargestNumber(nums, k)); // Output: "3"
}
}Python's sort method is used with a key function that considers the length of the string and its lexicographical order. The reverse parameter is set to True for descending order.
Time Complexity: O(n log n) due to sorting.
Space Complexity: O(1) auxiliary space, aside from the input storage.
Approach 2: Min-Heap (Priority Queue) Approach
Min-Heap Approach: This approach involves using a min-heap (or priority queue) to efficiently find the kth largest element. The idea is to maintain a heap of size k while iterating through the array of strings. By doing so, the smallest element in the heap will automatically become the kth largest once all numbers are processed. This method provides an optimal solution in terms of time complexity compared to sorting.
import heapq
def kthLargestNumber(nums, k):
minHeap = []
for num in nums:
heapq.heappush(minHeap, (len(num), num))
if len(minHeap) > k:
heapq.heappop(minHeap)
return minHeap[0][1]
print(kthLargestNumber(["3", "6", "7", "10"], 4)) # Output: "3"import java.util.*;
public class KthLargestNumber {
public static String kthLargestNumber(String[] nums, int k) {
PriorityQueue<String> minHeap = new PriorityQueue<>((a, b) -> {
if (a.length() != b.length()) return Integer.compare(a.length(), b.length());
return a.compareTo(b);
});
for (String num : nums) {
minHeap.offer(num);
if (minHeap.size() > k) minHeap.poll();
}
return minHeap.peek();
}
public static void main(String[] args) {
String[] nums = {"3", "6", "7", "10"};
int k = 4;
System.out.println(kthLargestNumber(nums, k)); // Output: "3"
}
}#include <iostream>
#include <vector>
#include <string>
#include <queue>
using namespace std;
string kthLargestNumber(vector<string>& nums, int k) {
auto comp = [](const string& a, const string& b) {
if (a.size() != b.size()) return a.size() > b.size();
return a > b;
};
priority_queue<string, vector<string>, decltype(comp)> minHeap(comp);
for (const string& num : nums) {
minHeap.push(num);
if (minHeap.size() > k) minHeap.pop();
}
return minHeap.top();
}
int main() {
vector<string> nums = {"3", "6", "7", "10"};
int k = 4;
cout << kthLargestNumber(nums, k) << endl; // Output: "3"
return 0;
}function kthLargestNumber(nums, k) {
const minHeap = [];
const pushHeap = (num) => {
minHeap.push(num);
let i = minHeap.length - 1;
while (i > 0 && compare(minHeap[i], minHeap[Math.floor((i - 1) / 2)]) < 0) {
[minHeap[i], minHeap[Math.floor((i - 1) / 2)]] = [minHeap[Math.floor((i - 1) / 2)], minHeap[i]];
i = Math.floor((i - 1) / 2);
}
};
const popHeap = () => {
const result = minHeap[0];
minHeap[0] = minHeap.pop();
let i = 0;
while (true) {
let left = 2 * i + 1, right = 2 * i + 2, smallest = i;
if (left < minHeap.length && compare(minHeap[left], minHeap[smallest]) < 0) smallest = left;
if (right < minHeap.length && compare(minHeap[right], minHeap[smallest]) < 0) smallest = right;
if (smallest !== i) {
[minHeap[i], minHeap[smallest]] = [minHeap[smallest], minHeap[i]];
i = smallest;
} else break;
}
return result;
};
const compare = (a, b) => {
if (a.length !== b.length) return a.length - b.length;
return a > b ? 1 : -1;
};
for (const num of nums) {
pushHeap(num);
if (minHeap.length > k) popHeap();
}
return minHeap[0];
}
console.log(kthLargestNumber(["3", "6", "7", "10"], 4)); // Output: "3";In Python, the heapq module is utilized for a min-heap to keep track of the k largest numbers. Tuple sorting by length and lexicographical value is employed, thus ensuring efficient management of heap size.
Time Complexity: O(n log k), since heap operations run in logarithmic time relative to k.
Space Complexity: O(k), which represents the heap's maintained size.
Approach 3: Sorting or Quickselect
We can sort the strings in the nums array in descending order as integers, and then take the k-th element. Alternatively, we can use the quickselect algorithm to find the k-th largest integer.
The time complexity is O(n times log n) or O(n), where n is the length of the nums array. The space complexity is O(log n) or O(1).
func kthLargestNumber(nums []string, k int) string {
sort.Slice(nums, func(i, j int) bool {
a, b := nums[i], nums[j]
if len(a) == len(b) {
return a > b
}
return len(a) > len(b)
})
return nums[k-1]
}Complexity comparison
| Approach | Complexity |
|---|---|
| Sort and Pick Approach | Time Complexity: O(n log n) due to sorting. |
| Min-Heap (Priority Queue) Approach | Time Complexity: O(n log k), since heap operations run in logarithmic time relative to k. |
| Sorting or Quickselect | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Sort and Pick | O(n log n) | O(1)–O(n) | Best when simplicity matters and full sorting cost is acceptable |
| Min-Heap (Priority Queue) | O(n log k) | O(k) | Efficient when k is small compared to n |
| Quickselect | Average O(n), Worst O(n²) | O(1) | Useful when you only need the kth element without sorting the entire array |
Frequently asked questions
Is Find the Kth Largest Integer in the Array easy or hard?
Find the Kth Largest Integer in the Array is classified as a Medium problem. The main challenge comes from handling numbers stored as strings and choosing an efficient selection method such as a heap or Quickselect instead of brute-force comparisons.
Find the Kth Largest Integer in the Array Python/Java solution
In Python, the problem can be solved using sorting with a custom key based on string length and lexicographic order, or by maintaining a heap using heapq. In Java, you can use Arrays.sort with a comparator or a PriorityQueue to maintain the top k elements.
How to solve Find the Kth Largest Integer in the Array in O(n)?
An average O(n) solution uses the Quickselect algorithm. The array is partitioned around a pivot, similar to quicksort, and only the side containing the kth largest element is processed recursively. Comparisons must be done using string length first and lexicographic order for equal lengths.
What is the best approach for Find the Kth Largest Integer in the Array?
The min-heap (priority queue) approach is typically the best balance between simplicity and efficiency. It keeps only the k largest elements in a heap of size k while iterating through the array. This results in O(n log k) time complexity and O(k) space, which is more efficient than sorting when k is much smaller than n.
Is Find the Kth Largest Integer in the Array asked at Google/Amazon/Meta?
Kth element and top‑k style problems frequently appear in interviews at companies like Amazon, Google, and Meta. Variants involving heaps, Quickselect, or sorting are common because they test understanding of selection algorithms and data structures such as priority queues.
What data structure is used in Find the Kth Largest Integer in the Array?
The most common data structure used is a min-heap (priority queue). It maintains the k largest elements seen so far while scanning the array. Sorting algorithms and Quickselect also rely on array manipulation and custom comparison logic for numeric strings.
What is the time complexity of Find the Kth Largest Integer in the Array?
Time complexity depends on the chosen algorithm. Sorting the array takes O(n log n). A min-heap solution improves this to O(n log k). A Quickselect-based selection algorithm can achieve average O(n) time but may degrade to O(n²) in the worst case.