Count three-digit numbers having difference X with its reverse
Given an integer X, the task is to count the total number of three-digit numbers having difference X with its reverse. If no such number exists, then print -1.
Input: X = 792
Output : 10
901 – 109 = 792
911 – 119 = 792
921 – 129 = 792
931 – 139 = 792
941 – 149 = 792
951 – 159 = 792
961 – 169 = 792
971 – 179 = 792
981 – 189 = 792
991 – 199 = 792
Input: X = 0
Approach: The given problem can be solved based on the following observations:
Let N = rpq
Therefore, N = 100r + 10q + p
Therefore, reverse of N = 100p + 10q + r
Therefore, the problem reduces to solving (100r + 10q + p) – (r + 10q + 100p) = X
-> 99(r – p) = X
-> r – p = X / 99
Therefore, if given X is a multiple of 99, then solution exists.
Follow the steps below to solve the problem based on the above observations:
- Check if X is multiple of 99 or not. If not found to be true, print -1 as no solution exists.
- Otherwise, calculate X / 99. Generate all pairs using digits [1, 9] and for each pair, check if their difference is equal to X / 99 or not.
- If found to be true for any pair, increase count by 10, as the middle digit can be permuted to place any value from the range [0, 9] for the obtained pair.
- Finally, print the value of the count obtained.
Below is the implementation of the above approach:
Time Complexity: O(1)
Auxiliary Space: O(1)
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