Data Structures - Arrays

1) Move all Negative numbers to the beginning of the Array
Move all the Negative elements to the beginning of the given Array. The solution is to take two pointers slowPointer, fastPointer. SlowPointer starting at position 0 and fastPointer starting at position 1. The required condition is to have all negatives at the beginning, so keep slowPointer at beginning as long as the element is positive and move it only incase the element is negative. For the fastPointer move it as long as the number is negative and stop when the number is positive and do a comparison with the slowPointer, if slowPointer is positive and fastPointer is negative swap the elements at both the positions and move both the pointers. When the Iteration is completed all the negative elements get moved to the beginning and all the positive elements to the end.

	private static void moveNegatives(int[]nums) {
	if(nums==null || nums.length==0) {
	return;
	}
	int i = 0;
	int j = 1;
	while(i<nums.length-1 && j<nums.length-1) {
	if(nums[j]<0 && nums[i]>=0) {
	int temp = nums[i];
	nums[i] = nums[j];
	nums[j] = temp;
	i++;
	j++;
	} else if(nums[i]<0) {
	i++;
	} else if(nums[j]>0) {
	j++;
	}
	}
	}

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Rotate an Image represented as a 2D matrix in place To rotate an image represented as a 2D Matrix in place we first need to transpose the matrix(move row to column and column to row), after the transpose is completed then reverse the positions of elements for each row which makes the complete 2D array rotation by 90 degrees.

	class Solution {
	public void rotate(int[][] matrix) {
	if(matrix==null || matrix.length==0) {
	return;
	}
	int n = matrix.length;
	for(int i=0;i<n;i++) {
	for(int j=i;j<n;j++) {
	int temp = matrix[i][j];
	matrix[i][j] = matrix[j][i];
	matrix[j][i] = temp;
	}
	}
	for(int i=0;i<n;i++) {
	for(int j=0;j<n/2;j++) {
	int temp = matrix[i][j];
	matrix[i][j] = matrix[i][n-1-j];
	matrix[i][n-1-j] = temp;
	}
	}
	}
	}

view raw Rotate Image represented as 2D Matrix hosted with ❤ by GitHub