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| 1 | +// 1st question using sliding window technique |
| 2 | +// 1st |
| 3 | +// MAXIMUM SUM SUB ARRAY OF SIZE K : using sliding window technique |
| 4 | +class GFG { |
| 5 | + // Returns maximum sum in |
| 6 | + // a subarray of size k. |
| 7 | + static int maxSum(int arr[], int n, int k) |
| 8 | + { |
| 9 | + // Initialize result |
| 10 | + int max_sum = Integer.MIN_VALUE; |
| 11 | + |
| 12 | + // Consider all blocks starting with i. |
| 13 | + for (int i = 0; i < n - k + 1; i++) { |
| 14 | + int current_sum = 0; |
| 15 | + for (int j = 0; j < k; j++) |
| 16 | + current_sum = current_sum + arr[i + j]; |
| 17 | + |
| 18 | + // Update result if required. |
| 19 | + max_sum = Math.max(current_sum, max_sum); |
| 20 | + } |
| 21 | + |
| 22 | + return max_sum; |
| 23 | + } |
| 24 | + |
| 25 | + // Driver code |
| 26 | + public static void main(String[] args) |
| 27 | + { |
| 28 | + int arr[] = { 1, 4, 2, 10, 2, 3, 1, 0, 20 }; |
| 29 | + int k = 4; |
| 30 | + int n = arr.length; |
| 31 | + System.out.println(maxSum(arr, n, k)); |
| 32 | + } |
| 33 | +} |
| 34 | + |
| 35 | +// 2nd question using sliding window technique |
| 36 | +// 2nd |
| 37 | +// Longest Substring Without Repeating Characters: using sliding window technique |
| 38 | +import java.util.*; |
| 39 | + |
| 40 | +public class GfG { |
| 41 | + |
| 42 | + // Function to find the length of the longest |
| 43 | + // substring without repeating characters |
| 44 | + static int longestUniqueSubstr(String s) { |
| 45 | + int n = s.length(); |
| 46 | + int res = 0; |
| 47 | + |
| 48 | + for (int i = 0; i < n; i++) { |
| 49 | + |
| 50 | + // Initializing all characters as not visited |
| 51 | + boolean[] visited = new boolean[256]; |
| 52 | + |
| 53 | + for (int j = i; j < n; j++) { |
| 54 | + |
| 55 | + // If current character is visited, |
| 56 | + // Break the loop |
| 57 | + if (visited[s.charAt(j)]) { |
| 58 | + break; |
| 59 | + } |
| 60 | + else { |
| 61 | + |
| 62 | + // Else update the result if this |
| 63 | + // window is larger, and mark current |
| 64 | + // character as visited. |
| 65 | + res = Math.max(res, j - i + 1); |
| 66 | + visited[s.charAt(j)] = true; |
| 67 | + } |
| 68 | + } |
| 69 | + } |
| 70 | + return res; |
| 71 | + } |
| 72 | + |
| 73 | + public static void main(String[] args) { |
| 74 | + String s = "geeksforgeeks"; |
| 75 | + System.out.println(longestUniqueSubstr(s)); |
| 76 | + } |
| 77 | +} |
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