```
int Solution::solve(int A, int B) {
// A -> no of eggs B-> no of total floors in building
//int dp[A+1][B+1];
//memset(dp, 0, sizeof(dp));
vector<vector<int>> dp(A+1,vector<int>(B+1,0));
// EXPLANATION: n->floors e->eggs count
// 1. F[n][e] = min(max(F[n-x][e], F[x-1][e-1]) + 1)
// x=1->n-1
// minimize the max(worst case) to get least no of tries. First term is
// when egg doesnt break. e remains same. We need to focus on higher floors (n-x)
// Second term is when egg breaks(e-1 eggs left). So to find breakpoint floor we go
// down (x-1). The 3rd term(+1) is for one try(dropping egg) done on current floor
// 2 constrains: F[1][e] = 1. one floor left and any no of eggs -> only 1 try
// F[n][1] = n. n floors and one egg left. Start from bottom floor and keep dropping
//same egg in each floor till it breaks (worst case N floors)
for(int i = 1; i<=B; i++) dp[i][1] = i; // 2nd constraint as seen above
for(int i = 1; i<=A; i++) dp[1][i] = 1; // 1st constraint
for(int i = 2; i<=B; i++){ // building
for(int j = 2; j<=A; j++){ // egg
dp[i][j] = INT_MAX;
// x = 1 to n-1
for(int x = 1; x<i; x++){
int brokenEggResult = dp[x-1][j-1];
int unbrokenEggResult = dp[i-x][j];
int temp = max(brokenEggResult, unbrokenEggResult) + 1;
if(temp < dp[i][j]) dp[i][j] = temp;
}
}
}
// its a bottoms up approach. So largest cell (bottom most rightmost val) is the ans
return dp[A][B];
}
```

If you find out why I’m getting the error, so let me know in the comments…