How do you find the number of Subarrays?

The number of all possible subarrays of an array of size N is N * (N + 1)/2.

How do you find an odd sum?

The total of any set of sequential odd numbers beginning with 1 is always equal to the square of the number of digits, added together. If 1,3,5,7,9,11,…, (2n-1) are the odd numbers, then; Sum of first odd number = 1. Sum of first two odd numbers = 1 + 3 = 4 (4 = 2 x 2).

How do you find the sum of all Subarrays?

Algorithm:

1. Traverse the array from start to end.
2. From every index start another loop from i to the end of array to get all subarray starting from i, keep a variable sum to calculate the sum.
3. For every index in inner loop update sum = sum + array[j]
4. If the sum is equal to the given sum then print the subarray.

How do I find all the subarrays of an array?

Approach:

1. Use three nested loops.
2. Outer loops will decide the starting point of a sub-array, call it as startPoint.
3. First inner loops will decide the group size (sub-array size).
4. The most inner loop will actually print the sub-array by iterating the given array from startPoint and print the next grps elements.

Where is all Subarrays in size k?

Sum of all subarrays of size K

1. Input: arr[] = {1, 2, 3, 4, 5, 6}, K = 3.
2. Output: 6 9 12 15.
3. Explanation: All subarrays of size k and their sum: Subarray 1: {1, 2, 3} = 1 + 2 + 3 = 6. Subarray 2: {2, 3, 4} = 2 + 3 + 4 = 9. Subarray 3: {3, 4, 5} = 3 + 4 + 5 = 12. Subarray 4: {4, 5, 6} = 4 + 5 + 6 = 15.

How many Subarrays are in an array Java?

A set of size ‘n’ can have a total of [n*(n+1)] / 2 subsets. Let the array be [1,2,3]. Subarrays are contiguous elements i.e 1,2,3,12,23,123 here. For n =3 , total number of subsets / subarrays are 6 ( [3* 4] / 2).

How do you find the number of odd numbers?

If N is odd,

1. If L or R is odd, then the count of odd number will be N/2 + 1 and even numbers = N – countofOdd.
2. Else, count of odd numbers will be N/2 and even numbers = N – countofOdd.

How do I generate Subarrays in size k?

How to find the number of subarrays in an array?

Another efficient approach is to first find the number of subarrays starting at index 0 and having an odd sum. Then traverse the array and update the number of subarrays starting at index i and having an odd sum. Below is the implementation of the above approach : Attention reader!

How do you count continuous odd numbers in an array?

Thus, for every K continuous odd numbers in the array, the count of sub-arrays with the odd product increases by K* (K+1)/2. One way of counting continuous odd numbers is to calculate the difference between the indexes of every two consecutive even numbers and subtract it by 1.

How do you find the odd product of an array?

Naive Approach: A simple solution is to calculate the product of every sub-array and check whether it is odd or not and calculate the count accordingly. Efficient Approach: An odd product is possible only by the product of odd numbers.