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LeetCode-494-Target-Sum.java
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LeetCode-494-Target-Sum.java
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/*
1.DFS
2.DP (Pull)
3.DP (Push)
4.DP (convert to subset problem & Pull)
5.DP (convert to subset problem & Push)
*/
class Solution {
// 1.DFS
/*
Time O(2^N)
*/
// public int findTargetSumWays(int[] nums, int S) {
// if (nums == null || nums.length == 0) return 0;
// int[] res = new int[1];
// DFS(nums, 0, S, res);
// return res[0];
// }
// private void DFS(int[] nums, int idx, int S, int[] res) {
// if (idx == nums.length) {
// if (S == 0) res[0]++;
// return;
// }
// DFS(nums, idx + 1, S + nums[idx], res);
// DFS(nums, idx + 1, S - nums[idx], res);
// }
// 2. DP (Knacpsack problem) (pull from previous level)
/*
this is a classic knapsack problem
in knapsack, we decide whether we choose this element or not
in this question, we decide whether we add this element or minus it
So start with a two dimensional array dp[i][j] which means the number of ways for first i-th element to reach a sum j
we can easily observe that dp[i][j] = dp[i-1][j+nums[i]] + dp[i-1][j-nums[i],
Another part which is quite confusing is return value, here we return dp[sum+S], why is that?
because dp's range starts from -sum --> 0 --> +sum
so we need to add sum first, then the total starts from 0, then we add S
Actually most of Sum problems can be treated as knapsack problem, hope it helps
*/
// public int findTargetSumWays(int[] nums, int S) {
// if (nums == null || nums.length == 0) return 0;
// int n = nums.length;
// int sum = 0;
// for (int num : nums) sum += num;
// if (S < -sum || S > sum) return 0;
// /*
// DP (pull)
// sub problem:
// int[][] state = new int[n + 1][2 * sum + 1];
// - state[i][j] means number of ways to sum to j using nums[0-i]
// but the j could range from [-sum, sum], so total possible sum is 2*sum+1
// sum value: -sum - 0 - sum
// index value: 0 - sum - 2*sum
// init:
// state[0][sum] = 1; // i==0 means it's empty set, so the sum value is 0, and we don't choose anything
// output:
// state[n][sum+S]
// recurrence relation:
// state[i][j] += state[i - 1][j + nums[i - 1]] if (j+nums[i - 1]) <= 2*sum
// state[i][j] += state[i - 1][j - nums[i - 1]] if (j-nums[i - 1]) >= to 0
// Time: O(N*sum) - N is number of elements, sum is the sum of elements
// Space: O(N*sum)
// */
// int[][] state = new int[n + 1][2 * sum + 1];
// // Init the state[0][sum] to be 1. sum means total value is 0, so if we select nothing, this is the only approach to get total value to be 1
// state[0][sum] = 1;
// for (int i = 1; i <= n; i++) {
// for (int j = 0; j < 2 * sum + 1; j++) {
// if (j + nums[i - 1] <= 2 * sum) state[i][j] += state[i - 1][j + nums[i - 1]];
// if (j - nums[i - 1] >= 0) state[i][j] += state[i - 1][j - nums[i - 1]];
// }
// }
// return state[n][sum + S];
// }
// DP (push to next level)
// public int findTargetSumWays(int[] nums, int S) {
// if (nums == null || nums.length == 0) return 0;
// int n = nums.length;
// int sum = 0;
// for (int num : nums) sum += num;
// if (S < -sum || S > sum) return 0;
// /*
// DP (pull)
// sub problem:
// state[i][j] means number of ways to sum to j using nums[0-i]
// sum value: -sum - 0 - sum
// index value: 0 - sum - 2*sum
// init:
// state[0][sum] = 1;
// output:
// state[n][sum+S]
// recurrence relation:
// state[i][j] += state[i + 1][j + nums[i]] if (j+nums[i - 1]) <= 2*sum
// state[i][j] += state[i + 1][j - nums[i]] if (j-nums[i - 1]) >= to 0
// Time: O(N*sum) - N is number of elements, sum is the sum of elements
// Space: O(N*sum)
// */
// int[][] state = new int[n + 1][2 * sum + 1];
// // Init the state[0][sum] to be 1. sum means total value is 0, so if we select nothing, this is the only approach to get total value to be 1
// state[0][sum] = 1;
// for (int i = 0; i < n; i++) {
// for (int j = 0; j < 2 * sum + 1; j++) {
// if (j + nums[i] <= 2 * sum) state[i + 1][j] += state[i][j + nums[i]];
// if (j - nums[i] >= 0) state[i + 1][j] += state[i][j - nums[i]];
// }
// }
// return state[n][sum + S];
// }
// 3.Optimized DP for space. Using one dimention array
/*
Time O(N*sum)
Space O(sum)
*/
// public int findTargetSumWays(int[] nums, int S) {
// if (nums == null || nums.length == 0) return 0;
// int n = nums.length;
// int sum = 0;
// for (int num : nums) sum += num;
// if (S < -sum || S > sum) return 0;
// int[] dp = new int[2 * sum + 1];
// dp[sum] = 1;
// for (int i = 1; i <= n; i++) {
// int[] tempDp = new int[2 * sum + 1];
// for (int j = 0; j <= 2 * sum; j++) {
// if (j + nums[i - 1] <= 2 * sum) {
// tempDp[j] += dp[j + nums[i - 1]];
// }
// if (j - nums[i - 1] >= 0) {
// tempDp[j] += dp[j - nums[i - 1]];
// }
// }
// dp = tempDp;
// }
// return dp[sum + S];
// }
// 4. DP (subset & pull & 2D array)
/*
Split the array to be two parts,
Let P denotes a subset of nums have a + in front of it
Let N denotes a subset of nums have a - in front of it
P ∪ N = {a1, a2, ..., an}, P ∩ N = ∅
sum(P) - sum(N) = target
sum(P) - sum(N) + sum(P) + sum(N) = target + sum(P) + sum(N)
2 * sum(p) = target + sum(P) + sum(N) <- Knapsack problem (find a subset has sum value equals to a special target)
subproblem:
dp[i][j] - the number of ways to nums[0 - i] elements or its subset to get sum to be j
recurrence relation:
dp[i][j] = dp[i-1][j] + dp[i-1][j - nums[i]] if j - nums[i] >= 0
init:
dp[0][sum] = 1
ans:
S = (S + sum) / 2 // update the new target
dp[n][S]
*/
// public int findTargetSumWays(int[] nums, int S) {
// if (nums == null || nums.length == 0) return 0;
// int n = nums.length;
// int sum = 0;
// for (int num : nums) sum += num;
// if (S < -sum || S > sum) return 0;
// if ((S + sum) % 2 != 0) return 0; // means we cannot fnid a subset P, which has 2 * sum(P) = target + sum, here target + sum must be even
// S = (S + sum) / 2;
// int[][] dp = new int[n + 1][sum + 1]; // means using [0, i] elements or its subset to get a sum to be j
// dp[0][0] = 1;
// for (int i = 1; i <= n; i++) {
// for (int j = 0; j <= sum; j++) {
// dp[i][j] = dp[i - 1][j];
// if (j - nums[i - 1] >= 0) {
// dp[i][j] += dp[i - 1][j - nums[i - 1]];
// }
// }
// // System.out.println(Arrays.deepToString(dp));
// }
// return dp[n][S];
// }
// DP: Subset & Optimized to use One-Dimension array & Pull (One of the best solution)
// public int findTargetSumWays(int[] nums, int S) {
// if (nums == null || nums.length == 0) return 0;
// int n = nums.length;
// int sum = 0;
// for (int num : nums) sum += num;
// if (S < -sum || S > sum) return 0;
// if ((S + sum) % 2 != 0) return 0; // means we cannot fnid a subset P, which has 2 * sum(P) = target + sum, here target + sum must be even
// S = (S + sum) / 2;
// // subproblem:
// int[] dp = new int[sum + 1];
// // init
// dp[0] = 1;
// // recurrence relation
// for (int i = 1; i <= n; i++) {
// for (int j = sum; j >= nums[i - 1]; j--) {
// if (j - nums[i - 1] >= 0) {
// dp[j] += dp[j - nums[i - 1]];
// }
// }
// }
// // res
// return dp[S];
// }
// DP (Pull && Subset problem && Optimized the Space)
public int findTargetSumWays(int[] nums, int S) {
if (nums == null || nums.length == 0) return 0;
int n = nums.length;
int sum = 0;
for (int num : nums) sum += num;
if (S < -sum || S > sum) return 0;
if ((S + sum) % 2 != 0) return 0; // means we cannot fnid a subset P, which has 2 * sum(P) = target + sum, here target + sum must be even
S = (S + sum) / 2;
// subproblem:
int[] dp = new int[S + 1]; // Optimized here, only need S
// init
dp[0] = 1;
// recurrence relation
for (int i = 1; i <= n; i++) {
for (int j = S; j >= nums[i - 1]; j--) {
if (j - nums[i - 1] >= 0) {
dp[j] += dp[j - nums[i - 1]];
}
}
}
// res
return dp[S];
}
// 5. DP (subset & push & 2D array)
/*
subproblem:
dp[i][j] - the sum of nums [0, i] or its subset is equal to j
recurrence relation:
dp[i + 1][j] += dp[i][j]
dp[i + 1][j + nums[i + 1]] += dp[i][j] // didn't select nums[i]
*/
// public int findTargetSumWays(int[] nums, int S) {
// if (nums == null || nums.length == 0) return 0;
// int n = nums.length;
// int sum = 0;
// for (int num : nums) sum += num;
// if (S < -sum || S > sum) return 0;
// if ((S + sum) % 2 != 0) return 0;
// S = (S + sum) / 2; // update new specific target
// // subproblem: the sum of nums [0, i] or its subset is equal to j
// int[][] dp = new int[n + 1][sum + 1];
// // init
// dp[0][0] = 1;
// // recurrence relation
// for (int i = 0; i < n; i++) {
// for (int j = 0; j <= sum; j++) {
// dp[i + 1][j] += dp[i][j];
// if (j + nums[i] <= sum) {
// dp[i + 1][j + nums[i]] += dp[i][j];
// }
// }
// }
// // ans
// return dp[n][S];
// }
}