2017-10-05 21:56:35 +08:00
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//Dynamic Programming solution for the Egg Dropping Puzzle
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public class EggDropping
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{
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// min trials with n eggs and m floors
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private static int minTrials(int n, int m)
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{
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int eggFloor[][] = new int[n+1][m+1];
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int result, x;
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for (int i = 1; i <= n; i++)
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{
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eggFloor[i][0] = 0; // Zero trial for zero floor.
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2017-10-05 22:02:03 +08:00
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eggFloor[i][1] = 1; // One trial for one floor
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2017-10-05 21:56:35 +08:00
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}
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2017-10-05 22:02:03 +08:00
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// j trials for only 1 egg
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2017-10-05 21:56:35 +08:00
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2017-10-05 22:02:03 +08:00
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for (int j = 1; j <= m; j++)
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2017-10-05 21:56:35 +08:00
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eggFloor[1][j] = j;
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// Using bottom-up approach in DP
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2017-10-05 22:02:03 +08:00
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for (int i = 2; i <= n; i++)
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{
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2017-10-05 21:56:35 +08:00
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for (int j = 2; j <= m; j++)
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{
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eggFloor[i][j] = Integer.MAX_VALUE;
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for (x = 1; x <= j; x++)
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{
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result = 1 + Math.max(eggFloor[i-1][x-1], eggFloor[i][j-x]);
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//choose min of all values for particular x
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2017-10-05 22:02:03 +08:00
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if (result < eggFloor[i][j])
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eggFloor[i][j] = result;
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2017-10-05 21:56:35 +08:00
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}
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}
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}
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2017-10-05 22:02:03 +08:00
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2017-10-05 21:56:35 +08:00
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return eggFloor[n][m];
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}
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//testing program
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public static void main(String args[])
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{
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2017-10-05 22:02:03 +08:00
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int n = 2, m = 4;
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//result outputs min no. of trials in worst case for n eggs and m floors
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int result = minTrials(n, m);
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System.out.println(result);
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2017-10-05 21:56:35 +08:00
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}
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}
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