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Kth Smallest Element

Certification: Advanced Level Accuracy: 0% Submissions: 0 Points: 15

Write a C++ program to find the kth smallest element in an n x n matrix where each row and column is sorted in ascending order.

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].
      • 
      
    • The 8th smallest number is 13.
      • 
    
    •
Example 2
    
  
  • Input: matrix = [
  [1,  4,  7, 11, 15],
  [2,  5,  8, 12, 19],
  [3,  6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
], k = 5
    • 
  
  • Output: 5
    • 
  
  • Explanation: 
      
      
    • The elements in the matrix are [1, 2, 3, 4, 5, ...].
      • 
      
    • The 5th smallest number is 5.
      • 
    
    •
Constraints
    
  
  • n == matrix.length
    • 
  
  • n == matrix[i].length
    • 
  
  • 1 ≤ n ≤ 300
    • 
  
  • -10^9 ≤ matrix[i][j] ≤ 10^9
    • 
  
  • All rows and columns in the matrix are sorted in ascending order
    • 
  
  • 1 ≤ k ≤ n^2
    • 
  
  • Time Complexity: O(n * log(max-min))
    • 
  
  • Space Complexity: O(1)
    •
ArraysAlgorithmsDeloitteSamsung
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Hint
Solution

Solution Hints

    
  
  • Use binary search to find the kth smallest element
    • 
  
  • For each mid value, count how many elements are less than or equal to it
    • 
  
  • Utilize the sorted property of rows and columns to efficiently count
    • 
  
  • You can also consider using a priority queue (min-heap) approach
    • 
  
  • Be careful about duplicate elements in the matrix
    •

Steps to solve by this approach:

 Step 1: Set the search range for binary search - start with the smallest element (top-left) and end with the largest element (bottom-right).
 Step 2: For each mid value in the binary search, count how many elements in the matrix are less than or equal to it.
 Step 3: To count efficiently, start from the bottom-left corner and move either right or up.
 Step 4: If the current count is less than k, move the left boundary up; otherwise, move the right boundary down.
 Step 5: Continue the binary search until left meets right.
 Step 6: The final value of left is the kth smallest element.
 Step 7: Be careful with duplicate values in the matrix - this approach handles them correctly.

Submitted Code :