# Python3 program to implement
# the above approach
# Function to return Mth element of
# array after k right rotations
def getFirstElement(a, N, K, M):
# The array comes to original state
# after N rotations
K %= N
# If K is greater or equal to M
if (K >= M):
# Mth element after k right
# rotations is (N-K)+(M-1) th
# element of the array
index = (N - K) + (M - 1)
# Otherwise
else:
# (M - K - 1) th element
# of the array
index = (M - K - 1)
result = a[index]
# Return the result
return result
# Driver Code
if __name__ == "__main__":
a = [ 1, 2, 3, 4, 5 ]
N = len(a)
K , M = 3, 2
print( getFirstElement(a, N, K, M))
# This code is contributed by chitranayal
Output
4
Time Complexity: O(1)
Auxiliary Space: O(1)
Please refer complete article on Mth element after K Right Rotations of an Array for more details!