Merge pull request #21 from vim12345/main

Add Some New Problem

Author: devvsakib

L-I/0001ReplacediagonalelementsineachrowofgivenMarixbyKthsmallestelementofthatrow/Readme.md

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+Question - Replace diagonal elements in each row of given Matrix by Kth smallest element of that row
+
+---------------------------------------------------------------------------------------------------------------
+
+Approach: The solution is based on the concept of sorting. Follow the steps mentioned below:
+
+    Traverse the matrix row-wise.
+    Copy this row in another list.
+    Sort the list and get the Kth smallest element and replace the diagonal element with that.
+
+Below is the implementation of the above approach.
+
+----------------------------------------------------------------------------------------------------------------
+Given a matrix mat[ ][ ] of size N*N and an integer K, containing integer values, the task is to replace diagonal elements by the Kth smallest element of row.
+
+Examples: 
+
+    Input: mat[][]= {{1, 2, 3, 4}
+                             {4, 2, 7, 6}
+                             {3, 5, 1, 9}
+                             {2, 4, 6, 8}}
+    K = 2
+    Output: 2, 2, 3, 4
+                 4, 4, 7, 6
+                 3, 5, 3, 8
+                 2, 4, 6, 4
+    Explanation: 2nd smallest element of 1st row = 2
+    2nd smallest element of 2nd row is 4
+    2nd smallest element of 3rd row is 3
+    2nd smallest element of 4th row is 4             
+
+    Input: mat[][] = {{1, 2, 3}
+                              {7, 9, 8}
+                              {2, 3, 6}}
+    K = 2
+    Output: 2, 2, 3
+                 7, 8, 8
+                 2, 3, 3

L-I/0001ReplacediagonalelementsineachrowofgivenMarixbyKthsmallestelementofthatrow/solution.js

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+	// JavaScript code for the above approach
+	let N = 4;
+
+	// Function to print Matrix
+	function showMatrix(mat) {
+		let i, j;
+		for (i = 0; i < N; i++) {
+			for (j = 0; j < N; j++) {
+				document.write(mat[i][j] + " ");
+			}
+			document.write('<br>')
+		}
+	}
+
+	// Function to return k'th smallest element
+	// in a given array
+	function kthSmallest(arr, n, K)
+	{
+		
+		// Sort the given array
+		arr.sort(function (a, b) { return a - b })
+
+		// Return k'th element
+		// in the sorted array
+		return arr[K - 1];
+	}
+
+	// Function to replace diagonal elements
+	// with Kth min element of row.
+	function ReplaceDiagonal(mat, K)
+	{
+		let i, j;
+		let arr = new Array(N);
+
+		for (i = 0; i < N; i++) {
+			for (j = 0; j < N; j++)
+				arr[j] = mat[i][j];
+			mat[i][i] = kthSmallest(arr, N, K);
+		}
+		showMatrix(mat);
+	}
+
+	// Utility Main function.
+	let mat = [[1, 2, 3, 4],
+	[4, 2, 7, 6],
+	[3, 5, 1, 9],
+	[2, 4, 6, 8]];
+
+	let K = 3;
+	ReplaceDiagonal(mat, K);
+

L-I/0003Findmaximumelementofeachrowinamatrix/Readme.md

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+Given a matrix, the task is to find the maximum element of each row.
+Examples: 
+
+Input :  [1, 2, 3]
+         [1, 4, 9]
+         [76, 34, 21]
+
+Output :
+3
+9
+76
+
+Input : [1, 2, 3, 21]
+        [12, 1, 65, 9]
+        [1, 56, 34, 2]
+Output :
+21
+65
+56

L-I/0003Findmaximumelementofeachrowinamatrix/solution.js

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+
+function maxelement(no_of_rows, arr)
+{
+	var i = 0;
+	
+	// Initialize max to 0 at beginning
+	// of finding max element of each row
+	var max = 0;
+	var result = Array.from({length: no_of_rows}, (_, i) => 0);
+	while (i < no_of_rows)
+	{
+		for (var j = 0; j < arr[i].length; j++)
+		{
+			if (arr[i][j] > max)
+			{
+				max = arr[i][j];
+			}
+		}
+		result[i] = max;
+		max = 0;
+		i++;
+
+	}
+	printArray(result);
+
+}
+
+// Print array element
+function printArray(result)
+{
+	for (var i = 0; i < result.length; i++)
+	{
+		document.write(result[i]+"<br>");
+	}
+
+}
+
+// Driver code
+	var arr = [[3, 4, 1, 8],
+	[ 1, 4, 9, 11],
+	[ 76, 34, 21, 1],
+[ 2, 1, 4, 5] ];
+	
+// Calling the function
+maxelement(4, arr);
+