Special Positions in a Binary Matrix | 2 Approaches | Leetcode-1582
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This is the 8th Video of our Playlist "Leetcode Easy".
In this video we will try to solve an easy but good practice problem - Special Positions in a Binary Matrix (Leetcode - 1582).
I will explain the intuition so easily that you will never forget and start seeing this as cakewalk EASYYY.
We will do live coding after explanation and see if we are able to pass all the test cases.
Also, please note that my Github solution link below contains both C++ as well as JAVA code.
Problem Name : Special Positions in a Binary Matrix
Company Tags : Will soon update
My solutions on Github(C++ & JAVA) : https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/Arrays/2-D%20Array/Leetcode%20Easy/Special%20Positions%20in%20a%20Binary%20Matrix.cpp
Leetcode Link : https://leetcode.com/problems/special-positions-in-a-binary-matrix/
My DP Concepts Playlist : https://youtu.be/7eLMOE1jnls
My Graph Concepts Playlist : https://youtu.be/5JGiZnr6B5w
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Approach Summary - Brute Force :
The method iterates through each element in the matrix, and for each non-zero element, it checks whether it is the only one in its row and column. If so, the counter variable result is incremented. The method then returns the total count of special elements in the matrix.The approach employs nested loops to traverse the matrix and validate each element's special status by comparing it with other elements in its row and column. The use of the boolean variable good helps determine if the current element is special based on the row and column checks.
Approach Summary - Better Force :
The approach here is different from the previous one. Instead of checking each element individually, this code first calculates the counts of 1s in each row and column separately using two additional vectors (rowCount and colCount). It then iterates through the matrix again, and for each 1 encountered, it checks whether its corresponding row and column have a count of 1. If so, the element is considered special, and the result counter is incremented. This approach is more efficient as it reduces the number of nested loops needed to validate each element, utilizing precomputed counts to determine special elements in a more straightforward manner.
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✨ Timelines✨
00:00 - Introduction
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