Sorting a 2D Array according to values in any given column in Java
We are given a 2D array of order N X M and a column number K ( 1<=K<=m). Our task is to sort the 2D array according to values in the Column K.
Examples:
Input : If our 2D array is given as (Order 4X4)
39 27 11 42
10 93 91 90
54 78 56 89
24 64 20 65
Sorting it by values in column 3
Output : 39 27 11 42
24 64 20 65
54 78 56 89
10 93 91 90
The Idea is to use Arrays.sort in Java.
// Java Code to sort 2D Matrix// according to any Columnimport java.util.*;class sort2DMatrixbycolumn { // Function to sort by column public static void sortbyColumn(int arr[][], int col) { // Using built-in sort function Arrays.sort Arrays.sort(arr, new Comparator<int[]>() { @Override // Compare values according to columns public int compare(final int[] entry1, final int[] entry2) { // To sort in descending order revert // the '>' Operator if (entry1[col] > entry2[col]) return 1; else return -1; } }); // End of function call sort(). } // Driver Code public static void main(String args[]) { int matrix[][] = { { 39, 27, 11, 42 }, { 10, 93, 91, 90 }, { 54, 78, 56, 89 }, { 24, 64, 20, 65 } }; // Sort this matrix by 3rd Column int col = 3; sortbyColumn(matrix, col - 1); // Display the sorted Matrix for (int i = 0; i < matrix.length; i++) { for (int j = 0; j < matrix[i].length; j++) System.out.print(matrix[i][j] + " "); System.out.println(); } }} |
Output:
39 27 11 42 24 64 20 65 54 78 56 89 10 93 91 90
Time complexity : O(n Log n) where n is number of rows. Here assumption is that the sort() function uses a O(n Log n) sorting algorithm.
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