Given N numbers, find the number of permutations in which the sum of elements at odd index and sum of elements at even index are equal.
Input: 1 2 3
The permutations are:
1 3 2 sum at odd index = 1+2 = 3, sum at even index = 3
2 3 1 sum at odd index = 2+1 = 3, sum at even index = 3
Input: 1 2 1 2
The permutations are:
1 2 2 1
2 1 1 2
2 2 1 1
The approach to the problem will be to use next_permutation() in C++ STL which helps to generate all the possible permutation of N numbers. If the sum of the odd index elements is equal to the sum of even index elements of the generated permutation, then increase the count. When all permutations are checked, print the count.
Below is the implementation of the above approach:
Time Complexity: O(N! * N)
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