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# Program to find lexicographically smallest string after applying operations in Python

Suppose we have a string s with only numeric digits and also have two values a and b. We can apply any one of the following two operations any number of times and in any order on s −

Add 'a' to all odd positioned items of s(0-indexed). If digit is 9, then by adding something with it will be cycled back to 0.

Rotate 's' to the right by b positions.

So we have to find the lexicographically smallest string we can get by applying the above operations any number of times on s.

So, if the input is like s = "5323" a = 9 b = 2, then the output will be 2050 because if we follow

- Rotate: "5323"
- Add: "5222"
- Add: "5121"
- Rotate: "2151"
- Add: "2050"

To solve this, we will follow these steps −

- seen := a new set
- deq := a new queue with one element 's'
- while deq is not empty, do
- curr := first deleted element of deq
- insert curr into seen set
- ad := perform given add operation on curr
- if ad is not in seen, then
- insert ad at the end of deq
- insert ad into seen set

- ro := perform rotate operation on curr
- if ro is not in seen, then
- insert ro at the end of deq
- insert ro into seen set

- return minimum of seen

## Example

Let us see the following implementation to get better understanding −

from collections import deque def add_(s,a): res = '' for idx, i in enumerate(s): if idx % 2 == 1: num = (int(i) + a) % 10 res += str(num) else: res += i return res def rotate_(s, b): idx = len(s)-b res = s[idx:] + s[0:idx] return res def solve(s, a, b): seen = set() deq = deque([s]) while deq: curr = deq.popleft() seen.add(curr) ad = add_(curr, a) if ad not in seen: deq.append(ad) seen.add(ad) ro = rotate_(curr, b) if ro not in seen: deq.append(ro) seen.add(ro) return min(seen) s = "5323" a = 9 b = 2 print(solve(s, a, b))

## Input

"5323", 9, 2

## Output

2050

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