Adding all the solutions

find-all-numbers-disappeared-in-an-array.java

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+// Time Complexity : O(n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : yes
+// Three line explanation of solution in plain english :  For each number in the array, mark its corresponding index as negative to indicate it exists. 
+// After marking, any index that remains positive represents a number that is missing. Collect all such missing numbers and return them.
+
+
+
+class Solution {
+    public List<Integer> findDisappearedNumbers(int[] nums) {
+        List<Integer> output = new ArrayList<>();
+
+        for(int i=0; i<nums.length; i++) {
+            int idx = Math.abs(nums[i]) -1;
+            if(nums[idx] > 0) {
+                nums[idx] *= -1;
+            }
+        }
+
+        for (int i=0; i<nums.length; i++) {
+            if(nums[i] > 0) {
+                output.add(i+1);
+            }
+
+        }
+        return output;
+    }
+}

game-of-life.java

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+// TimeComplexity: O(mn)
+// SpaceComplexity: O(1)
+// Explanation: We iterate through each cell, count its 8 neighbors, and mark state transitions in-place using temporary values (2 and 3) to preserve original states during computation. 
+// After the first pass, we convert the temporary states to their final values in a second pass.
+
+class Solution {
+    public void gameOfLife(int[][] board) {
+        // 1-->0 ==2;
+        // 0-->1 ==3;
+        int m = board.length;
+        int n = board[0].length;
+        for(int i=0; i<m; i++) {
+            for(int j=0; j<n; j++) {
+                int count = countAlive(board, i, j);
+                if (board[i][j] == 1 && (count < 2 || count > 3)) {
+                board[i][j] = 2;          // live → dead
+            } else if (board[i][j] == 0 && count == 3) {
+                board[i][j] = 3;          // dead → live
+            }
+
+            }   
+        }
+
+        for(int i=0; i<m; i++) {
+            for(int j=0; j<n; j++) {
+                if(board[i][j] == 2) {
+                    board[i][j] = 0;
+                } else if(board[i][j] == 3) {
+                    board[i][j] = 1;
+                }
+            }
+        }
+    }
+
+    private int countAlive(int[][] board, int i, int j) {
+
+         int count =0;
+         int[][] dir = {{-1, -1}, {-1, 0}, {-1, 1},{ 0, -1},{ 0, 1},{ 1, -1}, { 1, 0}, { 1, 1}};
+         for (int k =0; k<dir.length; k++) {
+            int newi = i+dir[k][0];
+            int newj = j+dir[k][1];
+            if(newi > -1 && newi < board.length && newj >-1 && newj < board[0].length && (board[newi][newj] == 1 || board[newi][newj] == 2)){
+                count+=1;
+            }
+         }
+
+         return count;
+    }
+}

maximum-and-minimum-in-an-array.java

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+// Time Complexity : O(n)
+// Space Complexity : O(1)
+// Three line explanation of solution in plain english : Initialize min and max using the first one or two elements depending on array size. 
+// Iterate through the rest of the array in pairs, updating min and max efficiently. Return a list containing the final minimum and maximum values.
+
+class Solution {
+    public static List<Integer> findMinMax(ArrayList<Integer> arr) {
+        List<Integer> result = new ArrayList<>();
+        int n = arr.size();
+        int mini, maxi, i;
+
+        if (n % 2 == 1) {
+            mini = maxi = arr.get(0);
+            i = 1;
+        } else {
+            if (arr.get(0) < arr.get(1)) {
+                mini = arr.get(0);
+                maxi = arr.get(1);
+            } else {
+                mini = arr.get(1);
+                maxi = arr.get(0);
+            }
+            i = 2;
+        }
+
+        // Process elements in pairs
+        while (i < n - 1) {
+            if (arr.get(i) < arr.get(i + 1)) {
+                mini = Math.min(mini, arr.get(i));
+                maxi = Math.max(maxi, arr.get(i + 1));
+            } else {
+                mini = Math.min(mini, arr.get(i + 1));
+                maxi = Math.max(maxi, arr.get(i));
+            }
+            i += 2;
+        }
+
+        result.add(mini);
+        result.add(maxi);
+        return result;
+    }
+}