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Kth Smallest Element in a Sorted Matrix

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 of the rows and columns are sorted in ascending order. The matrix elements are sorted both row-wise and column-wise.

Example 1
    
  
  • Input: matrix = [[1,5,9],[10,11,13],[12,13,15]], k = 8
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  • Output: 13
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  • Explanation: 
    • The elements in the matrix are: [1,5,9,10,11,12,13,13,15]. 
    • When sorted: [1,5,9,10,11,12,13,13,15].
    • The 8th smallest element (1-indexed) is 13.
    •
Example 2
    
  
  • Input: matrix = [[-5]], k = 1
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  • Output: -5
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  • Explanation: 
    • The matrix contains only one element: -5. 
    • The 1st smallest element is -5. 
    • Therefore, the answer is -5.
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Constraints
    
  
  • n == matrix.length == matrix[i].length
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  • 1 <= n <= 300
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  • -10^9 <= matrix[i][j] <= 10^9
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  • All the rows and columns of matrix are guaranteed to be sorted in non-decreasing order
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  • 1 <= k <= n^2
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  • Time Complexity: O(n * log(max - min))
    • 
  
  • Space Complexity: O(1)
    •
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Hint
Solution

Solution Hints

    
  
  • Use binary search on the range of values in the matrix
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  • The smallest value is matrix[0][0] and largest is matrix[n-1][n-1]
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  • For each mid value, count how many elements are <= mid
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  • Count elements efficiently by starting from top-right or bottom-left corner
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  • If count >= k, search in lower half; otherwise search in upper half
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  • Use the property that rows and columns are sorted
    •

Steps to solve by this approach:

 Step 1: Initialize left pointer to matrix[0][0] (smallest element) and right pointer to matrix[n-1][n-1] (largest element)
 Step 2: Use binary search on the value range, calculate mid = left + (right - left) / 2
 Step 3: Count how many elements in matrix are <= mid using efficient counting from bottom-left corner
 Step 4: If count < k, the kth element is larger, so move left = mid + 1
 Step 5: If count >= k, the kth element is <= mid, so move right = mid
 Step 6: Continue until left == right, which gives us the kth smallest element
 Step 7: Return the final value as the kth smallest element

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