Leetcode Magical String problem solution YASH PAL, 31 July 2024 In this Leetcode Magical String problem solution Given an integer n, return the largest palindromic integer that can be represented as the product of two n-digits integers. Since the answer can be very large, return it modulo 1337. Problem solution in Python. class Solution(object): def magicalString(self, n): """ :type n: int :rtype: int """ q = collections.deque() mask = 1 ^ 2 cnt, num, res = 0, 1, 0 for _ in range(n): q.append(num) res += num == 1 cnt += 1 if cnt == q[0]: q.popleft() cnt = 0 num ^= mask return res Problem solution in Java. class Solution { public int magicalString(int n) { int[] A = new int[n+2]; int fast=1; int slow=1; int num= 1; while(fast<=n){ A[fast++] = num; if (A[slow++]==2){ A[fast++] = num; } num = 3-num; } int count=0; for(int j=1;j<=n;j++){ if(A[j]==1) count++; } return count; } } Problem solution in C++. class Solution { public: int magicalString(int n) { if(n<=0) return 0; // has 0 1's if(n<=3) return 1; // has one 1's int count =0; vector<int> _string; _string.push_back(1); _string.push_back(2); _string.push_back(2); int idx = 2; // the last digit index , decides how many digits will be added to the end of the string int digit = 1; while(_string.size()<n){ if(_string[idx]==1){ _string.push_back(digit); } if(_string[idx]==2){ _string.push_back(digit); _string.push_back(digit); } if(digit==1) digit = 2; else digit =1; idx++; } for(int x = 0; x<n; x++){ if(_string[x]==1) count++; // Counting only till the n specified in the question } return count; } }; coding problems