Solution 1

Python
class Solution:
    def isPossibleToRearrange(self, s: str, t: str, k: int) -> bool:
        eqLen = len(s)//k
        sCounter = defaultdict(int)
        for i in range(0, len(s), eqLen):
            sCounter[s[i:i+eqLen]] += 1
            
        for i in range(0, len(t), eqLen):
            part = t[i:i+eqLen]
            if sCounter[part] < 1:
                return False
            sCounter[part] -= 1
        return True
Leet Code/python.py · L3513–3526