package com.freetymekiyan.algorithms.level.medium;

import org.junit.After;
import org.junit.Assert;
import org.junit.Before;
import org.junit.Test;

import java.util.Arrays;

/**
 * Given an integer array with all positive numbers and no duplicates, find the number of possible combinations that add
 * up to a positive integer target.
 * <p>
 * Example:
 * <p>
 * nums = [1, 2, 3]
 * target = 4
 * <p>
 * The possible combination ways are:
 * (1, 1, 1, 1)
 * (1, 1, 2)
 * (1, 2, 1)
 * (1, 3)
 * (2, 1, 1)
 * (2, 2)
 * (3, 1)
 * <p>
 * Note that different sequences are counted as different combinations.
 * Therefore the output is 7.
 * <p>
 * Follow up:
 * What if negative numbers are allowed in the given array?
 * How does it change the problem?
 * What limitation we need to add to the question to allow negative numbers?
 * <p>
 * Company Tags: Google, Snapchat, Facebook
 * Tags: Dynamic Programming
 * Similar Problems: (M) Combination Sum
 * <p>
 * Answer to follow up:
 * 1. The recurrence relation still works. But base case breaks.
 * 2. Each number is only used one time. Think about [1, -1], 1.
 */
public class CombinationSum4 {

    private CombinationSum4 c;

    private int[] dp;

    /**
     * DP. Bottom-up.
     * Recurrence relation:
     * comb[target] = sum(comb[target - nums[i]]), where 0 <= i < nums.length, and target >= nums[i].
     * Base case:
     * comb[0] = 1. Think about [1], 1, where comb[1] = comb[1 - 1] = comb[0] = 1.
     */
    public int combinationSum4(int[] nums, int target) {
        int[] comb = new int[target + 1];
        comb[0] = 1;
        for (int i = 1; i < comb.length; i++) {
            for (int j = 0; j < nums.length; j++) {
                if (i - nums[j] >= 0) { // Array's not sorted. Need to check each number.
                    comb[i] += comb[i - nums[j]];
                }
            }
        }
        return comb[target];
    }

    /**
     * DP. Top-down.
     */
    public int combinationSum4TopDown(int[] nums, int target) {
        dp = new int[target + 1];
        Arrays.fill(dp, -1);
        dp[0] = 1;
        return helper(nums, target);
    }

    private int helper(int[] nums, int target) {
        if (target == 0) {
            return 1;
        }
        if (dp[target] != -1) {
            return dp[target];
        }
        int res = 0;
        for (int i = 0; i < nums.length; i++) {
            if (target >= nums[i]) {
                res += helper(nums, target - nums[i]);
            }
        }
        dp[target] = res;
        return res;
    }

    @Before
    public void setUp() {
        c = new CombinationSum4();
    }

    @Test
    public void testExamples() {
        int[] nums = {1, 2, 3};
        int target = 4;
        Assert.assertEquals(7, c.combinationSum4(nums, target));
        target = 6;
        Assert.assertEquals(24, c.combinationSum4(nums, target));
    }

    @After
    public void tearDown() {
        c = null;
    }

}

@@ -1225,16 +1225,85 @@
 tom-up.%0A
+     * State: S%5Bi%5D means the # of combinations that can reach sum i.%0A
      * R
@@ -1305,26 +1305,25 @@
   * Recurren
-ce
+t
  relation:%0A 
@@ -1332,46 +1332,30 @@
   * 
-comb%5Btarget%5D = sum(comb%5Btarget
+S%5Bi%5D = sum(S%5Bi
  - nums%5B
 i%5D%5D)
@@ -1350,17 +1350,17 @@
  - nums%5B
-i
+j
 %5D%5D), whe
@@ -1367,17 +1367,17 @@
 re 0 %3C= 
-i
+j
  %3C nums.
@@ -1403,17 +1403,17 @@
 %3E= nums%5B
-i
+j
 %5D.%0A     
@@ -1432,20 +1432,17 @@
 %0A     * 
-comb
+S
 %5B0%5D = 1.
@@ -1472,22 +1472,16 @@
 ere 
-comb
+S
 %5B1%5D = 
-comb
+S
 %5B1 -
@@ -1486,20 +1486,17 @@
  - 1%5D = 
-comb
+S
 %5B0%5D = 1.

update combination sum 4