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refactor: BoardPath
(#5431)
refactor: BoardPath Co-authored-by: alxkm <alx@alx.com>
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@ -1,46 +1,16 @@
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package com.thealgorithms.dynamicprogramming;
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/*
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* this is an important Algo in which
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* we have starting and ending of board and we have to reach
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* we have to count no. of ways
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* that help to reach end point i.e number by rolling dice
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* which have 1 to 6 digits
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Test Case:
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here target is 10
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int n=10;
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startAlgo();
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System.out.println(bpR(0,n));
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System.out.println(endAlgo()+"ms");
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int[] strg=new int [n+1];
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startAlgo();
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System.out.println(bpRS(0,n,strg));
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System.out.println(endAlgo()+"ms");
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startAlgo();
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System.out.println(bpIS(0,n,strg));
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System.out.println(endAlgo()+"ms");
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*/
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public final class BoardPath {
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private BoardPath() {
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}
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public static long startTime;
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public static long endTime;
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public static void startAlgo() {
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startTime = System.currentTimeMillis();
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}
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public static long endAlgo() {
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endTime = System.currentTimeMillis();
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return endTime - startTime;
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}
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/**
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* Recursive solution without memoization
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*
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* @param start - the current position
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* @param end - the target position
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* @return the number of ways to reach the end from the start
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*/
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public static int bpR(int start, int end) {
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if (start == end) {
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return 1;
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@ -54,6 +24,14 @@ public final class BoardPath {
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return count;
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}
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/**
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* Recursive solution with memoization
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*
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* @param curr - the current position
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* @param end - the target position
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* @param strg - memoization array
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* @return the number of ways to reach the end from the start
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*/
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public static int bpRS(int curr, int end, int[] strg) {
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if (curr == end) {
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return 1;
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@ -71,15 +49,23 @@ public final class BoardPath {
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return count;
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}
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/**
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* Iterative solution with tabulation
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*
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* @param curr - the current position (always starts from 0)
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* @param end - the target position
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* @param strg - memoization array
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* @return the number of ways to reach the end from the start
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*/
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public static int bpIS(int curr, int end, int[] strg) {
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strg[end] = 1;
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for (int i = end - 1; i >= 0; i--) {
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int count = 0;
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for (int dice = 1; dice <= 6 && dice + i < strg.length; dice++) {
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for (int dice = 1; dice <= 6 && dice + i <= end; dice++) {
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count += strg[i + dice];
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}
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strg[i] = count;
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}
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return strg[0];
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return strg[curr];
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}
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}
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