Count pairs in array such that one element is reverse of another | Function to reverse the digits of the number ; Loop to iterate till the number is greater than 0 ; Extract the last digit and keep multiplying it by 10 to get the reverse of the number ; Function to find the pairs from the such that one number is reverse of the other ; Iterate through all pairs ; Increment count if one is the reverse of other ; Driver code
def reverse ( num ) : NEW_LINE INDENT rev_num = 0 NEW_LINE while ( num > 0 ) : NEW_LINE INDENT rev_num = rev_num * 10 + num % 10 NEW_LINE num = num // 10 NEW_LINE DEDENT return rev_num NEW_LINE DEDENT def countReverse ( arr , n ) : NEW_LINE INDENT res = 0 NEW_LINE for i in range ( n ) : NEW_LINE INDENT for j in range ( i + 1 , n ) : NEW_LINE INDENT if ( reverse ( arr [ i ] ) == arr [ j ] ) : NEW_LINE INDENT res += 1 NEW_LINE DEDENT DEDENT DEDENT return res NEW_LINE DEDENT if __name__ == ' _ _ main _ _ ' : NEW_LINE INDENT a = [ 16 , 61 , 12 , 21 , 25 ] NEW_LINE n = len ( a ) NEW_LINE print ( countReverse ( a , n ) ) NEW_LINE DEDENT