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DP_SticklerThief.java
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DP_SticklerThief.java
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import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
// https://practice.geeksforgeeks.org/problems/stickler-theif/0
// https://leetcode.com/problems/house-robber/
// https://practice.geeksforgeeks.org/problems/max-sum-without-adjacents/0 (Similar Problem)
// https://www.geeksforgeeks.org/maximum-sum-such-that-no-two-elements-are-adjacent/
// https://www.geeksforgeeks.org/find-maximum-possible-stolen-value-houses/
// https://leetcode.com/problems/house-robber/discuss/156523/From-good-to-great.-How-to-approach-most-of-DP-problems. (BEST Resource)
// https://www.youtube.com/watch?v=xlvhyfcoQa4 (Good video explanation)
class DP_SticklerThief {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while(t-- > 0) {
int n = Integer.parseInt(br.readLine());
String[] s = br.readLine().trim().split("\\s+");
int[] ip = new int[n];
for (int i = 0; i < n; i++) {
ip[i] = Integer.parseInt(s[i]);
}
//System.out.println(getMaximumMoney1(ip));
//System.out.println(getMaximumMoney2(ip));
System.out.println(getMaximumMoney3(ip)); // BEST approach. Easy to understand.
}
}
private static int getMaximumMoney1(int[] ip) {
int include = ip[0];
int exclude = 0;
for (int i = 1; i < ip.length; i++) {
int oldInclude = include;
include = Math.max(include, exclude+ip[i]);
exclude = Math.max(exclude, oldInclude);
}
return Math.max(include, exclude);
}
// APPROACH 2 : Using dynamic programming
// time complexity : O(n), space complexity : O(1)
private static int getMaximumMoney2(int[] house) {
int n = house.length;
if(n == 1) return house[0];
if(n == 2) return Math.max(house[0], house[1]);
int[] maxMoney = new int[n];
maxMoney[0] = house[0];
maxMoney[1] = Math.max(house[0], house[1]);
/* on every ith position,
* a robber has two options : 1. rob this current house 2. don't rob this house
* if he robs current ith house, then total money he gets is : (house[i] + maxMoney[i - 2])
* if he don't rob current ith house, then total money he gets is : maxMoney[i - 1]
* now we have to take the max between these two options on each ith position
* to get the maximum amount of money at the end of the array
* */
for (int i = 2; i < n; i++) {
maxMoney[i] = Math.max(house[i] + maxMoney[i - 2], maxMoney[i - 1]);
}
/*each step we choose the max money we can rob,
so at the last index we'll have the max money we can make from whole array*/
return maxMoney[n-1];
}
// APPROACH 2 : Using dynamic programming
// time complexity : O(n), space complexity : O(1)
private static int getMaximumMoney3(int[] house) {
int n = house.length;
if(n == 1) return house[0];
if(n == 2) return Math.max(house[0], house[1]);
int moneyRobbedFromSecondLastHouse = house[0];
int moneyRobbedFromLastHouse = Math.max(house[0], house[1]);
int moneyRobbedFromCurrentHouse = 0;
/* We can notice that in the method 2 we use only maxMoney[i-2] and maxMoney[i-1],
* so going just 2 steps back. We can hold them in 2 variables instead.
* This optimization is met in Fibonacci sequence creation
* */
for (int i = 2; i < n; i++) {
moneyRobbedFromCurrentHouse = Math.max(house[i] + moneyRobbedFromSecondLastHouse, moneyRobbedFromLastHouse);
moneyRobbedFromSecondLastHouse = moneyRobbedFromLastHouse;
moneyRobbedFromLastHouse = moneyRobbedFromCurrentHouse;
}
return moneyRobbedFromCurrentHouse;
}
}