Find the missing number in 1 to N

Author: sathishmepco

InterviewPrograms/src/com/java/array/FindMissingNo.java

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+package com.java.array;
+/*
+ *	Find the Missing Number in 1 to N
+ *
+ * An array contains 1 to N numbers 
+ * but one number in the array is missing
+ * Write the java program to find the missing no
+ * 
+ * Steps:
+ * 1. Calculate the sum of the array
+ * 2. Calculate the sum of 1 to N
+ * 3. Subtract array from the expected sum
+ * 4. The result value is the missing no
+ * 
+ *  Formula for Calculate sum of 1 to N
+ *  1 + 2+ 3 + 4 + 5 + ...... + N
+ *  sum = (N * (N+1)) / 2
+ *  
+ *  array = {2, 4, 1, 6, 7, 5, 3, 9};
+ *  array Length = 8
+ *  sum of array = 37
+ *  N = array Length + 1 (missing no) = 9
+ *  sum of 1 to N =  45
+ *  missing no = sum of 1 to N - array sum
+ *  missing no = 45 - 37
+ * 
+ * Note::
+ * This approach will work only for array contains
+ * value from 1 to n (order does not matter)
+ */
+public class FindMissingNo {
+	public static void main(String[] args) {
+//		int array[] = {2, 4, 1, 6, 7, 5, 3, 9};
+		int array[] = {2, 4, 1, 6, 3};
+		int n = array.length;
+
+		//calculate the sum of array
+		int arraySum = 0;
+		for(int v : array)
+			arraySum += v;
+
+		//array length is 8
+		//one no is missing, then n should by n+1
+		// here 8+1
+		n = n+1;
+		
+		//this calculates the sum from 1 to (n).
+		//1+2+3+....n
+		int expectedSum = (n * (n+1)) / 2;
+		
+		System.out.println("The Given Array is : ");
+		for(int v : array)
+			System.out.print(v+" ");
+		
+		int missingNo = expectedSum - arraySum;
+		System.out.println("\nMissing Number is : "+missingNo);
+	}
+}
+/*
+	OUTPUT
+	
+	The Given Array is : 
+	2 4 1 6 7 5 3 9 
+	Missing Number is : 8
+	
+	The Given Array is : 
+	2 4 1 6 3 
+	Missing Number is : 5
+*/