CB Singh

Solution approach : We will be using sliding window. Deque will be used to store useful element of current subarray. Add first element to deque. Then for adding next element, check if earlier elements are smaller, then remove all elements from deque. Maximum will always be at front of deque. For sliding and removing, check if arr[i] == front of queue, then remove the element. import java.util.Arrays ; import java.util.Deque ; import java.util.LinkedList ; public class MaxOfAllSubarrays { private static int [] findMaxOfAllSubArrays ( int [] arr , int subArraySize) { int [] answerArr = new int [arr. length - subArraySize + 1 ] ; int i = 0 , j = 0 , count = 0 ; Deque<Integer> deque = new LinkedList<>() ; while (j < arr. length ) { while (!deque.isEmpty() && deque.getLast() < arr[j]) deque.removeLast() ; deque.add(arr[j]) ; if (j - i + 1 == subArraySize) { answe...

We would be solving this problem using  Sliding Window approach with 1 Hashmap First we will populate pattern map with occurrence of chars in pattern map. Additionally, a new indicator distinctCharsInMapValueMoreThanZero will be created. It's initial value would be map.size(). This variable is for numbers of chars in map with value > 0. This will ensure we don't have to compare all values of map to see, if all characters are zero or not. When variable distinctCharsInMapValueMoreThanZero is zero, it means current substring is anagram of pattern.  When traversing string via Sliding window, we will keep reduce count of character in pattern map. In map, we will not push any char, which are not part of pattern.

We would be solving this problem using  Sliding Window approach with 1 Hashmap First we will populate pattern map with occurrence of chars in pattern map. Additionally, a new indicator distinctCharsInMapValueMoreThanZero will be created. It's initial value would be map.size(). This variable is for numbers of chars in map with value > 0. This will ensure we don't have to compare all values of map to see, if all characters are zero or not. When variable  distinctCharsInMapValueMoreThanZero is zero, it means current substring is anagram of pattern.  When traversing string via Sliding window, we will keep reduce count of character in pattern map. In map, we will not push any char, which are not part of pattern.

  There are 3 approach to solve this problem : Brute Force  Time complexity - O (nk) Using list to store element with sliding window approach Time complexity - O(n) Auxiliary space - O(k)  Using sliding window approach without list By traversing array from reverse. Time complexity - O(n)

 There are 2 approaches to handle this problem : Brute force way Optimized approach is via using Sliding Window. public class MaxSumSubArray {     //Driver program public static void main(String[] args) { int []arr = {2,3,5,2,9,7,1}; int subArraySize = 3; System.out.println(calcMaxSumSubArrayBruteForce(arr, subArraySize)); System.out.println(calcMaxSumSubArraySlidingWindow(arr, subArraySize)); } private static int calcMaxSumSubArrayBruteForce (int[] arr, int subArraySize) { int maxSum = 0; for (int i = 0; i <=arr.length-subArraySize ; i++) { int currentSum = 0; for (int j = i; j < i+subArraySize; j++) { currentSum += arr[j]; } if(currentSum > maxSum) maxSum = currentSum; } return maxSum; } private static int calcMaxSumSubArraySlidingWindow (int[] arr, int subArraySize) { int maxSum=Inte...