Competitive Coding - 1 : Find Missing Number in Array Completed

MissingNumberInArray.java

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+// Given a list of n-1 integers and these integers are in the range of 1 to n. 
+// There are no duplicates in list. One of the integers is missing in the list. 
+// Write an efficient code to find the missing integer. 
+// Input : arr[] = [1, 2, 3, 4, 6, 7, 8]
+// Output : 5
+
+
+// Input : arr[] = [1, 2, 3, 4, 5, 6, 8, 9]
+// Output : 7
+
+// Time Complexity : o(logn)
+// Space Complexity : o(1)
+// Approach : In the Array If any number is missing then difference between the number and the index will be greater than 1. Since it is a sorted Array,
+// Using Binary Search we can eliminate the portion of the array where the difference is not equal between mid - low and mid - high.
+// finally if the array containing 2 elements then missing  number = avg of two numbers
+
+public int Search(int nums[] , int size){
+    // [1, 2, 3, 4, 6, 7, 8]
+    // [0, 1, 2, 3, 4, 5, 6]
+    //size = 7
+    int low = 0; // 0
+    int high = size-1; // 7-1 =6
+    while(high-low>=2){ // loop executes if you have atleast three elements
+        int mid = low + (high-low)/2; //3 //1
+        // step 1 -  [1,2,3,4,6,7,8] nums
+        // step 2 - [4,6,7,8] -> [3,4,5,6] indexes
+        // step 3 - [4,6] -> [3,4] indexes
+        if(nums[mid]-mid != nums[low]-low){
+            // missing element in the left array 
+            high=mid;
+        } else if (nums[mid]-mid != nums[high]-high){
+            // missing element in the right array
+            low=mid;
+        }
+    }
+    return nums[low]+num[high] /2 ;
+}
+// Approach 2: Binary Search
+public int Search(int nums[] , int size){
+    // [1, 2, 3, 4, 6, 7, 8]
+    // [0, 1, 2, 3, 4, 5, 6]
+    //size = 7
+    int low = 0; // 0
+    int high = size-1; // 7-1 =6
+    while((high-low)>1){ // loop executes if you have atleast three elements
+        int mid = low + (high-low)/2; //3 //1
+        // step 1 -  [1,2,3,4,6,7,8] nums
+        // step 2 - [4,6,7,8] -> [3,4,5,6] indexes
+        // step 3 - [4,6] -> [3,4] indexes
+        if(nums[mid]-mid != nums[low]-low){
+            // missing element in the left
+            high=mid;
+        } else if (nums[mid]-mid != nums[high]-high){
+            // missing element in the right array
+            low=mid;
+        }
+    }
+    return nums[low]+1 ;
+}
+