What is the time complexity of Kth Smallest Element in a Sorted Matrix?

The optimal solution has a time complexity of O(n*log(m)).

What data structures are used to solve Kth Smallest Element in a Sorted Matrix?

This problem uses Array, Binary Search, Sorting, Heap (Priority Queue), Matrix. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Kth Smallest Element in a Sorted Matrix?

Insight 1: The matrix is sorted both row-wise and column-wise, which enables the use of binary search. Insight 2: Instead of directly searching for the k-th smallest element, we can use binary search to find a potential 'mid' value and count how many elements in the matrix are less than or equal to 'mid'.

Which companies ask LeetCode 378. Kth Smallest Element in a Sorted Matrix in interviews?

This problem is frequently asked at OKX, PhonePe, Salesforce, X.

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Kth Smallest Element in a Sorted Matrix - LeetCode 378 Solution

Kth Smallest Element in a Sorted Matrix - Complete Solution Guide

Kth Smallest Element in a Sorted Matrix is LeetCode problem 378, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an n x n matrix where each of the rows and columns is sorted in ascending order, return the k th smallest element in the matrix . Note that it is the k th smallest element in the sorted order , not the k th distinct element. You must find a solution with a memory complexity better than O(n 2 ) . Example 1: Input: matrix = [[1,5,9],[10,11,13],[12,13,15]], k = 8 Output: 13 Explanation: The elements in the matrix are [1,5,9,10,11,12,13, 13 ,15], and the 8 th smallest number is 13 Example 2: I

Detailed Explanation

The problem asks us to find the k-th smallest element in an n x n matrix where each row and column is sorted in ascending order. It is important to note that we are looking for the k-th smallest element in the *sorted order* of all elements in the matrix, not the k-th distinct element. The memory complexity should be better than O(n^2), hinting that we should avoid simply flattening the matrix into a single sorted array. We are given the matrix size 'n' and an integer 'k' as input. The output is the k-th smallest element in the matrix.

Solution Approach

The solution utilizes binary search. The search space is defined by the minimum and maximum values in the matrix (matrix[0][0] and matrix[n-1][n-1] respectively). For each 'mid' value during the binary search, we iterate through the matrix and count the number of elements that are less than or equal to 'mid'. This counting is done efficiently by starting from the top-right corner of each row and moving towards the left until we find an element that is less than or equal to 'mid'. The number of elements to the left of the current position (including the current position) gives the count of elements less than or equal to 'mid' in that row. The sum of these counts for all rows gives the total count of elements less than or equal to 'mid' in the matrix. Based on whether this count is less than or greater than or equal to 'k', the binary search adjusts its lower or upper bounds.

Step-by-Step Algorithm

Step 1: Initialize 'low' to the smallest element in the matrix (matrix[0][0]) and 'high' to the largest element in the matrix (matrix[n-1][n-1]).
Step 2: Perform binary search while 'low' is less than or equal to 'high'.
Step 3: Calculate the 'mid' value as 'low + (high - low) // 2'.
Step 4: Initialize 'count' to 0 and 'j' to 'n - 1'.
Step 5: Iterate through each row 'i' of the matrix.
Step 6: While 'j' is within bounds (>= 0) and matrix[i][j] is greater than 'mid', decrement 'j'.
Step 7: Add 'j + 1' to 'count' (since all elements from matrix[i][0] to matrix[i][j] are less than or equal to 'mid').
Step 8: After iterating through all rows, check if 'count' is less than 'k'.
Step 9: If 'count' is less than 'k', it means that the k-th smallest element is greater than 'mid', so update 'low' to 'mid + 1'.
Step 10: Otherwise, it means that the k-th smallest element is less than or equal to 'mid', so update 'high' to 'mid - 1'.
Step 11: After the binary search finishes (i.e., 'low' > 'high'), 'low' will be the k-th smallest element. Return 'low'.

Key Insights

Insight 1: The matrix is sorted both row-wise and column-wise, which enables the use of binary search.
Insight 2: Instead of directly searching for the k-th smallest element, we can use binary search to find a potential 'mid' value and count how many elements in the matrix are less than or equal to 'mid'.
Insight 3: If the count is less than k, the k-th smallest element must be greater than 'mid', so we adjust the lower bound of the binary search. Otherwise, the k-th smallest element is less than or equal to 'mid', so we adjust the upper bound.

Complexity Analysis

Time Complexity: O(n*log(m))

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search, Sorting, Heap (Priority Queue), Matrix.

Companies

Asked at: OKX, PhonePe, Salesforce, X.

Kth Smallest Element in a Sorted Matrix | LeetCode 378

Medium

Array Binary Search Sorting Heap (Priority Queue)Matrix

Time: O(n*log(m))Space: O(1)

Solution Code

Available in:

class Solution {
    public int kthSmallest(int[][] matrix, int k) {
        int n = matrix.length;
        int low = matrix[0][0];
        int high = matrix[n - 1][n - 1];

        while (low <= high) {
            int mid = low + (high - low) / 2;
            
            int count = 0;
            int j = n - 1;
            for (int i = 0; i < n; i++) {
                while (j >= 0 && matrix[i][j] > mid) {
                    j--;
                }
                count += (j + 1);
            }
            
            if (count < k) {
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        
        return low;
    }
}

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Problem Info

Description

Problem 378: Kth Smallest Element in a Sorted Matrix - Detailed explanation and solution approaches.

Topics

Array Binary Search Sorting Heap (Priority Queue)Matrix

Companies

OKXPhonePeSalesforceX

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:378

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

11 steps

1
Step 1: Initialize 'low' to the smallest element in the matrix (matrix[0][0]) and 'high' to the largest element in the matrix (matrix[n-1][n-1]).
2
Step 2: Perform binary search while 'low' is less than or equal to 'high'.
3
Step 3: Calculate the 'mid' value as 'low + (high - low) // 2'.
4
Step 4: Initialize 'count' to 0 and 'j' to 'n - 1'.
5
Step 5: Iterate through each row 'i' of the matrix.
6
Step 6: While 'j' is within bounds (>= 0) and matrix[i][j] is greater than 'mid', decrement 'j'.
7
Step 7: Add 'j + 1' to 'count' (since all elements from matrix[i][0] to matrix[i][j] are less than or equal to 'mid').
8
Step 8: After iterating through all rows, check if 'count' is less than 'k'.
9
Step 9: If 'count' is less than 'k', it means that the k-th smallest element is greater than 'mid', so update 'low' to 'mid + 1'.
10
Step 10: Otherwise, it means that the k-th smallest element is less than or equal to 'mid', so update 'high' to 'mid - 1'.
11
Step 11: After the binary search finishes (i.e., 'low' > 'high'), 'low' will be the k-th smallest element. Return 'low'.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*log(m))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The binary search algorithm utilizes constant extra space for variables like 'low', 'high', 'mid', 'count', and 'j'. Therefore the space complexity does not scale with the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Heap (Priority Queue)

Use a min-heap to store the elements. Initially, push the first element of each row into the heap. Then, extract the minimum element from the heap k times. After each extraction, push the next element from the same row into the heap.

Time: O(k * log(n)), where n is the size of the matrix. Each extraction and insertion into the heap takes O(log(n)) time, and we perform this k times.Space: O(n), to store the elements in the heap.

Trade-offs: The heap approach might be faster when k is significantly smaller than n^2, but has higher space complexity compared to the binary search approach. Also, this implementation is more complex compared to the binary search.

Sorting

Flatten the matrix into a single array, sort the array, and return the k-th element. This is simple to implement but does not meet the O(n^2) memory constraint asked by the prompt.

Time: O(n^2 * log(n)), where n is the dimension of matrix. Flattening takes O(n^2), sorting takes O(n^2 * log(n^2)) = O(n^2 * log(n)).Space: O(n^2), to store the flattened array.

Trade-offs: This approach is not space efficient and is only useful for small matrix sizes and is not a suitable method.

✨

Key Insights

1Insight 1: The matrix is sorted both row-wise and column-wise, which enables the use of binary search.
2Insight 2: Instead of directly searching for the k-th smallest element, we can use binary search to find a potential 'mid' value and count how many elements in the matrix are less than or equal to 'mid'.
3Insight 3: If the count is less than k, the k-th smallest element must be greater than 'mid', so we adjust the lower bound of the binary search. Otherwise, the k-th smallest element is less than or equal to 'mid', so we adjust the upper bound.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: k = 1. The algorithm correctly identifies the smallest element (matrix[0][0]) as the 1st smallest.
Edge case 2: k = n*n. The algorithm correctly identifies the largest element (matrix[n-1][n-1]) as the n*n-th smallest.
Edge case 3: All elements in the matrix are the same. The binary search will converge to that value, and the counting logic will also work correctly.
Edge case 4: Matrix contains negative numbers. The algorithm functions correctly as the binary search range is initialized to the minimum and maximum elements of the matrix, which could be negative.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly initializing the binary search range. The range should be from the smallest to the largest element in the matrix, not from 0 to some arbitrary large number.

✗Common mistake 2: Not handling the case where 'count' is equal to 'k' correctly. The binary search should continue until 'low' and 'high' converge to the k-th smallest element. Prematurely returning when count == k might result in incorrect answer.

✗Common mistake 3: Using integer division that rounds in the wrong direction, which can cause the binary search to get stuck in an infinite loop. Using `low + (high - low) // 2` is crucial to preventing this.

✗Common mistake 4: Not considering the matrix is sorted in both rows and columns. This will lead to inefficient brute-force solutions.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Finding the median salary in a company where employee salaries are grouped into different departments and each department has a sorted list of salaries. K would be set to n/2 if n is the total number of employees.

🏦Real-world scenario 2: In database systems, finding the k-th smallest value in a multi-dimensional index can be accelerated using this type of search if the index is maintained in a sorted fashion.

🎮Real-world scenario 3: Resource allocation problems where resources are ranked and need to be allocated based on a priority order. Finding the k-th most important resource is a similar problem.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 19, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Kth Smallest Element in a Sorted Matrix"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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