Given an array arr[] of size N, Convert the given array as even numbers in the array occurs at the odd index, odd numbers in the array occur at the even index, and whether the sum of the numbers in an array is divisible by the size of an array. If the array follows all the properties then an array is valid else it is invalid.
An array will be valid if it follows all the given three conditions:
- In the given array arr[], at every even index position, it must contain an odd number.
- In the given array arr[], at every odd index position, it must contain an even number.
- The sum of the given array arr[] must be divisible by the size of the given array.
Note: Array is 0 indexes based.
Examples:
Input : arr = {1, 2, 3, 4, 5, 6, 9, 10}
Output: "VALID"
Explanation:
1.Sum of given array is 40, and size of array is 8
2.At every even index position array containing odd number
3.At every odd position index array containing even number
Input : arr = {11, 4, 9, 3, 13}
Output: "INVALID"
Explanation:
1.Sum of given array is 40, and size of array is 4
2.At index 3 it containing an odd value which does not follow given condition
Hence it is invalid.
Approach:
- Traverse the given array and check for every element whether they satisfy the condition or not.
- Check for every index:
- if the index is odd, then the value at that index must be even.
- Else if the index is even, then the value at that index must be odd.
- Else print array is invalid and return.
- Store the sum of every given element.
- At last check it is divisible from the given array size or not.
- If the sum is divisible then print “VALID” else print “INVALID”.
Below is the implementation of the above approach:
Java
// Sum of Array Divisible by Size with Even// and Odd Numbers at Odd and Even Index // in Javapublic class Main { // check whether array is valid or not public static String validate(int[] arr) { // Finding size of given array int N = arr.length; // Store sum of given array int s = 0; // traverse the given array for (int i = 0; i < N; i++) { // store sum of elements s += arr[i]; // if index is even and value is // odd, then continue if (i % 2 == 0 && arr[i] % 2 == 1) continue; // if index is odd and value is // even, then continue else if (i % 2 == 1 && arr[i] % 2 == 0) continue; // violeting given condition else return "INVALID"; } // check last condition, sum is // divisible by size or not return s % N == 0 ? "VALID" : "INVALID"; } // Driver Code public static void main(String[] args) { int[] arr = { 1, 2, 3, 4, 5, 6, 9, 10 }; System.out.println(validate(arr)); int[] barr = { 11, 4, 9, 3, 13 }; System.out.println(validate(barr)); }} |
Output
VALID INVALID
- Time Complexity: O(N)
- Space Complexity: O(1)
