Done Array-2

problem1.java

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+// Time Complexity : O(n)
+// Space Complexity : O(1)  (output list not counted)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No, just need to use abs() while marking.
+
+
+// Your code here along with comments explaining your approach in three sentences only
+// Since values are from 1 to n, we use each value as an index and mark that index as visited by making it negative.
+// After marking, any position that still has a positive number means that (index+1) never appeared in the array.
+// Finally, we collect all such missing numbers into the answer list.
+
+import java.util.*;
+
+class Solution {
+    public List<Integer> findDisappearedNumbers(int[] nums) {
+
+        List<Integer> res = new ArrayList<>();
+
+        for (int i = 0; i < nums.length; i++) {
+            int idx = Math.abs(nums[i]) - 1;
+            if (nums[idx] > 0) nums[idx] = -nums[idx];
+        }
+
+        for (int i = 0; i < nums.length; i++) {
+            if (nums[i] > 0) res.add(i + 1);
+        }
+
+        return res;
+    }
+}

problem2.java

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+// Time Complexity : O(n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : Only thing is handling odd length array properly.
+
+
+// Your code here along with comments explaining your approach in three sentences only
+// Instead of comparing every element with both min and max, we compare elements in pairs.
+// For each pair, we first decide which one is smaller and which one is bigger, then update min and max accordingly.
+// This reduces comparisons and gives better than 2*(n-2) comparisons.
+
+class Solution {
+    public int[] findMinMax(int[] nums) {
+
+        int n = nums.length;
+        int min, max;
+        int i = 0;
+
+        // init min and max based on first one or first two
+        if (n % 2 == 0) {
+            if (nums[0] < nums[1]) {
+                min = nums[0];
+                max = nums[1];
+            } else {
+                min = nums[1];
+                max = nums[0];
+            }
+            i = 2;
+        } else {
+            min = nums[0];
+            max = nums[0];
+            i = 1;
+        }
+
+        // compare in pairs
+        while (i < n - 1) {
+
+            int a = nums[i];
+            int b = nums[i + 1];
+
+            if (a < b) {
+                if (a < min) min = a;
+                if (b > max) max = b;
+            } else {
+                if (b < min) min = b;
+                if (a > max) max = a;
+            }
+
+            i += 2;
+        }
+
+        return new int[]{min, max};
+    }
+}

problem3.java

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+// Time Complexity : O(m*n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : Need to carefully count live neighbors using old state only.
+
+
+// Your code here along with comments explaining your approach in three sentences only
+// We update the board in-place by using extra marker states so we don’t lose original info.
+// 2 means dead -> live (was 0, will be 1) and 3 means live -> dead (was 1, will be 0).
+// After first pass, we convert markers to final 0/1 values in second pass.
+
+class Solution {
+    public void gameOfLife(int[][] board) {
+
+        int m = board.length;
+        int n = board[0].length;
+
+        int[] dr = {-1,-1,-1,0,0,1,1,1};
+        int[] dc = {-1,0,1,-1,1,-1,0,1};
+
+        for (int r = 0; r < m; r++) {
+            for (int c = 0; c < n; c++) {
+
+                int live = 0;
+
+                // count live neighbors based on old values
+                for (int k = 0; k < 8; k++) {
+                    int nr = r + dr[k];
+                    int nc = c + dc[k];
+
+                    if (nr >= 0 && nr < m && nc >= 0 && nc < n) {
+                        // old live is 1 or 3 (3 means it was live before update)
+                        if (board[nr][nc] == 1 || board[nr][nc] == 3) {
+                            live++;
+                        }
+                    }
+                }
+
+                // apply rules with markers
+                if (board[r][c] == 1) {
+                    if (live < 2 || live > 3) board[r][c] = 3; // live -> dead
+                } else {
+                    if (live == 3) board[r][c] = 2; // dead -> live
+                }
+            }
+        }
+
+        // convert markers to final state
+        for (int r = 0; r < m; r++) {
+            for (int c = 0; c < n; c++) {
+                if (board[r][c] == 2) board[r][c] = 1;
+                else if (board[r][c] == 3) board[r][c] = 0;
+            }
+        }
+    }
+}