// min trials with n eggs and m floors

private static int minTrials(int n, int m) {

int[][] eggFloor = new int[n + 1][m + 1];

int result, x;

for (int i = 1; i <= n; i++) {

eggFloor[i][0] = 0; // Zero trial for zero floor.

eggFloor[i][1] = 1; // One trial for one floor

}

// j trials for only 1 egg

for (int j = 1; j <= m; j++)

eggFloor[1][j] = j;

// Using bottom-up approach in DP

for (int i = 2; i <= n; i++) {

for (int j = 2; j <= m; j++) {

eggFloor[i][j] = Integer.MAX_VALUE;

for (x = 1; x <= j; x++) {

result = 1 + Math.max(eggFloor[i - 1][x - 1], eggFloor[i][j - x]);

// choose min of all values for particular x

if (result < eggFloor[i][j])

eggFloor[i][j] = result;

}

}

}

return eggFloor[n][m];

}

public static void main(String args[]) {

int n = 2, m = 4;

// result outputs min no. of trials in worst case for n eggs and m floors

int result = minTrials(n, m);

System.out.println(result);

}