Optimised : 0448-find-all-numbers-disappeared-in-an-array

Author: jaiswalrohan8796

javascript/0448-find-all-numbers-disappeared-in-an-array.js

@@ -17,3 +17,31 @@ var findDisappearedNumbers = function(nums) {
 
     return Array.from(numberSet);
 };
+
+//Optimized Solution for Follow-up
+
+//time complexity O(n) , space complexity O(1)
+
+var findDisappearedNumbers = function(nums) {
+    for(let i = 0; i < nums.length; i++) {
+        let curr  = Math.abs(nums[i])
+        let idx = curr - 1
+        if(nums[idx] > 0) {
+            nums[idx] = nums[idx] * (-1)
+        }
+    }
+    let res = []
+    for(let i = 0; i < nums.length; i++) {
+        if(nums[i] > 0) {
+            res.push(i + 1)
+        }
+    }
+    return res
+};
+
+// For each value in the array mark its presence by making the number negative at that place in array
+// eg. if you hae array [3,1,4,1] for 3, i will go to index 2 and make its value negative ie. now nums[2] becomes -4. present array: [3,1,-4,1]
+// for 1, i will go to index 0 and make its value negative ie. now nums[0] becomes -3. present array: [-3,1,-4,1]
+// for 4, (take abs value), i will go to index 3 and make its value negative ie. now nums[3] becomes -1. present array: [-3,1,-4,-1]
+// for 1 take abs value), i will go to index 0 as it is already -ve do nothing. present array: [-3,1,-4,-1]
+// At last I will have [-3,1,-4,-1]. now i will iterate over the array, whichever idx has positive value that number will not be in the array so as we have nums[1]>0 so 2 is not in the list.