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LeetCode-1049-Last-Stone-Weight-II.java
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LeetCode-1049-Last-Stone-Weight-II.java
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/*
Similar to: https://leetcode.com/problems/target-sum/
这题实际上就是在每个element前面加 +，-，然后求和。如果最后的和是负，那我们可以等价转换成整数，比如：
[1,2,3]
[-1,2,-3], sum -2, could be convert to:
[1,-2,3], sum +2
所以我们就减少了一半的运算。
*/
class Solution {
// 1.DFS without optimization (TLE)
/*
Adding + or - to each number, and get the value to be minimum.
*/
// int min = Integer.MAX_VALUE;
// public int lastStoneWeightII(int[] stones) {
// int n = stones.length;
// /*
// 1 <= stones.length <= 30
// 1 <= stones[i] <= 100
// sum -3000 - 0 - 3000
// index 0 - 3000 - 6000
// subproblem:
// mem[i][j] - means number of sum answers, using the nums [0, i] and add +,- to the element, and get the number of different sum
// */
// DFS(stones, 0, 0);
// return min;
// }
// private void DFS(int[] stones, int idx, int sum) {
// if (idx == stones.length) {
// if (sum >= 0) {
// min = Math.min(min, sum);
// }
// return;
// }
// DFS(stones, idx + 1, sum + stones[idx]);
// DFS(stones, idx + 1, sum - stones[idx]);
// }
// 2.DFS with optimization (using memorized array)
/*
https://leetcode.com/problems/last-stone-weight-ii/discuss/313051/Simple-Java-DFS%2BMemoization-Solution-Knapsacktarget-sum
*/
// public int lastStoneWeightII(int[] stones) {
// int n = stones.length;
// /*
// 1 <= stones.length <= 30
// 1 <= stones[i] <= 100
// sum -3000 - 0 - 3000
// index 0 - 3000 - 6000
// subproblem:
// mem[i][j] - means the minimum lost weight of the nums [0, i] and add +,- to the elements
// */
// Integer[][] mem = new Integer[n + 1][6001];
// return DFS(stones, 0, 0, mem);
// }
// private int DFS(int[] stones, int idx, int sum, Integer[][] mem) {
// if (idx == stones.length) {
// /*
// the weight left must be negative
// [1,2,3]
// [-1,2,-3], sum -2, could be convert to:
// [1,-2,3], sum +2
// */
// return Math.abs(sum);
// }
// int t = sum + 3000; // making the sum + offset to get the real index
// if (mem[idx][t] != null) return mem[idx][t]; // optimization
// mem[idx][t] = Math.min(DFS(stones, idx + 1, sum + stones[idx], mem), DFS(stones, idx + 1, sum - stones[idx], mem));
// return mem[idx][t];
// }
// 3.DP (pull, my own solution)
/*
subproblem:
dp[i][j] - minimum weight by calculating the sum of nums [0, i], with adding +,- in front of each element
recurrence relation:
for (int i = 1; i <= n; i++) {
for (k = 0; k <= sum; k++) {
if (dp[i - 1][k] != Integer.MAX_VALUE) {
int sumPlus = Math.abs(dp[i - 1][k] + stones[k - 1]);
dp[i][sumPlus] = Math.min(dp[i][sumPlus], sumPlus);
int sumMinus = Math.abs(dp[i - 1][k] - stones[k - 1]);
dp[i][sumMinus] = Math.min(dp[i][sumMinus], sumMinus);
}
}
}
init
d[0][0-sum] = Integer.MAX_VALUE
ans
min positive (d[[0][0-sum]])
Time Complexity O(N*sum) - N is number of elements in array, sum is sum of all elements
Space: O(N*sum)
Runtime: 2 ms, faster than 73.85% of Java online submissions for Last Stone Weight II.
Memory Usage: 35.8 MB, less than 100.00% of Java online submissions for Last Stone Weight II.
*/
// public int lastStoneWeightII(int[] stones) {
// int n = stones.length;
// int sum = 0;
// for (int i : stones) sum += i;
// // subproblem: dp[i][j] means the minimum lost weight by calculating the sum of nums [0, i], with +,- in front of each element
// int[][] dp = new int[n + 1][sum + 1];
// // init
// for (int i = 1; i <= n; i++) {
// for (int j = 0; j <= sum; j++) {
// dp[i][j] = Integer.MAX_VALUE;
// }
// }
// dp[1][stones[0]] = stones[0];
// // recurrence relation
// for (int i = 2; i <= n; i++) {
// for (int j = 0; j <= sum; j++) {
// if (dp[i - 1][j] != Integer.MAX_VALUE) {
// int sumPlus = Math.abs(dp[i - 1][j] + stones[i - 1]);
// dp[i][sumPlus] = Math.min(dp[i][sumPlus], sumPlus);
// int sumMinus = Math.abs(dp[i - 1][j] - stones[i - 1]);
// dp[i][sumMinus] = Math.min(dp[i][sumMinus], sumMinus);
// }
// }
// }
// int min = Integer.MAX_VALUE;
// for (int j = 0; j <= sum; j++) {
// min = Math.min(min, dp[n][j]);
// }
// return min;
// }
// 3.DP (push, my own solution)
/*
Runtime: 2 ms, faster than 73.85% of Java online submissions for Last Stone Weight II.
Memory Usage: 35.7 MB, less than 100.00% of Java online submissions for Last Stone Weight II.
*/
// public int lastStoneWeightII(int[] stones) {
// int n = stones.length;
// int sum = 0;
// for (int i : stones) sum += i;
// // subproblem: dp[i][j] means the minimum lost weight by calculating the sum of stones [0, i], with +,- in front of each element
// int[][] dp = new int[n + 1][sum + 1];
// // initialize
// for (int i = 0; i <= n; i++) {
// for (int j = 0; j <= sum; j++) {
// dp[i][j] = Integer.MAX_VALUE;
// }
// }
// dp[1][stones[0]] = stones[0];
// // recurrence relation
// for (int i = 1; i <= n - 1; i++) {
// for (int j = 0; j <= sum; j++) {
// if (dp[i][j] != Integer.MAX_VALUE) {
// int sumPlus = Math.abs(dp[i][j] + stones[i]);
// dp[i + 1][sumPlus] = Math.min(dp[i + 1][sumPlus], sumPlus);
// int sumMinus = Math.abs(dp[i][j] - stones[i]);
// dp[i + 1][sumMinus] = Math.min(dp[i + 1][sumMinus], sumMinus);
// }
// }
// }
// int min = Integer.MAX_VALUE;
// for (int j = 0; j <= sum; j++) {
// min = Math.min(min, dp[n][j]);
// }
// return min;
// }
// DP (pull && Optimized using Boolean)
// public int lastStoneWeightII(int[] stones) {
// int n = stones.length;
// int sum = 0;
// for (int i : stones) sum += i;
// // subproblem: dp[i][j] means using the elements stones [0, i], with adding +,- in front of each element, we could sum up to j (true if we could, false if we could not)
// boolean[][] dp = new boolean[n + 1][sum + 1];
// // init
// dp[0][0] = true;
// // recurrence relation
// for (int i = 1; i <= n; i++) {
// for (int j = 0; j <= sum; j++) {
// if (dp[i - 1][j]) {
// int sumPlus = Math.abs(j + stones[i - 1]);
// dp[i][sumPlus] |= dp[i - 1][j];
// int sumMinus = Math.abs(j - stones[i - 1]);
// dp[i][sumMinus] |= dp[i - 1][j];
// }
// }
// }
// // ans
// for (int j = 0; j <= sum; j++) {
// if (dp[n][j]) return j;
// }
// return -1;
// }
// // DP: Pull & Optimized using boolean & Using one dimension array
// public int lastStoneWeightII(int[] stones) {
// int n = stones.length;
// int sum = 0;
// for (int i : stones) sum += i;
// // subproblem: dp[j] means using the elements stones [0, i], with adding +,- in front of each element, we could sum up to j (true if we could, false if we could not)
// boolean[] dp = new boolean[sum + 1];
// // init
// dp[0] = true;
// // recurrence relation
// for (int i = 1; i <= n; i++) {
// boolean[] temp = new boolean[sum + 1];
// for (int j = 0; j <= sum; j++) {
// if (dp[j]) {
// int sumPlus = Math.abs(j + stones[i - 1]);
// temp[sumPlus] |= dp[j];
// int sumMinus = Math.abs(j - stones[i - 1]);
// temp[sumMinus] |= dp[j];
// }
// }
// dp = temp;
// }
// // ans
// for (int j = 0; j <= sum; j++) {
// if (dp[j]) return j;
// }
// return -1;
// }
// // Push (Optimized using boolean)
// public int lastStoneWeightII(int[] stones) {
// int n = stones.length;
// int sum = 0;
// for (int i : stones) sum += i;
// // subproblem: dp[i][j] means using the elements stones [0, i], with adding +,- in front of each element, we could sum up to j (true if we could, false if we could not)
// boolean[][] dp = new boolean[n + 1][sum + 1];
// // init
// dp[0][0] = true;
// // recurrence relation
// for (int i = 0; i < n; i++) {
// for (int j = 0; j <= sum; j++) {
// if (dp[i][j]) {
// int sumPlus = Math.abs(j + stones[i]);
// dp[i + 1][sumPlus] = true;
// int sumMinus = Math.abs(j - stones[i]);
// dp[i + 1][sumMinus] = true;
// }
// }
// }
// // ans
// for (int j = 0; j <= sum; j++) {
// if (dp[n][j]) return j;
// }
// return -1;
// }
//Push (Optimized using boolean && using 1-d array)
// public int lastStoneWeightII(int[] stones) {
// int n = stones.length;
// int sum = 0;
// for (int i : stones) sum += i;
// // subproblem: dp[j] means using the elements stones [0, i], with adding +,- in front of each element, we could sum up to j (true if we could, false if we could not)
// boolean[] dp = new boolean[sum + 1];
// // init
// dp[0] = true;
// // recurrence relation
// for (int i = 0; i < n; i++) {
// boolean[] temp = new boolean[sum + 1];
// for (int j = 0; j <= sum; j++) {
// if (dp[j]) {
// int sumPlus = Math.abs(j + stones[i]);
// temp[sumPlus] = true;
// int sumMinus = Math.abs(j - stones[i]);
// temp[sumMinus] = true;
// }
// }
// dp = temp;
// }
// // ans
// for (int j = 0; j <= sum; j++) {
// if (dp[j]) return j;
// }
// return -1;
// }
// 4. DP (Convert to subset problem)
/*
Use P to denote a subset, all elements add + in front of the elements
Use N to denote a subset, all elements add - in front of the elements
sum(P) - sum(N) = target (target is the minimum abs of positive or negative value, as a mentioned before negative sum could be equally convert to positive sum)
sum(P) - sum(N) + sum(P) + sum(N) = target + sum(P) + sum(N)
2 * sum(P) = target + sum (target + sum must be an even number)
so target = 2 * sum(P) - sum, making target to be minimuze abs positive or negative number
We convert the problem to be find a subset, the sum is equal to (target + sum) / 2 to be minimum positive number. And the target is the returned value.
subproblem:
dp[i][j] - from the subset in nums [0,i], with adding +,- in front of each element, calculate the sum, which is j
recurrence relation:
dp[i][j] = dp[i - 1][j]
dp[i][j] |= dp[i - 1][j - nums[i - 1]] if j - nums[i - 1] >= 0
init:
dp[0][0] = true
ans:
find the j that makes |2*j - sum| minimum, and return the minimum result
*/
// public int lastStoneWeightII(int[] stones) {
// int n = stones.length;
// int sum = 0;
// for (int i : stones) sum += i;
// // subproblem: the sum of subset stones [0,j] with adding +,-, in front of each element, calculated sum is j
// boolean[][] dp = new boolean[n + 1][sum + 1];
// // init
// dp[0][0] = true;
// // recurrence relation
// for (int i = 1; i <= n; i++) {
// for (int j = 0; j <= sum; j++) {
// dp[i][j] = dp[i - 1][j];
// if (j - stones[i - 1] >= 0) {
// dp[i][j] |= dp[i - 1][j - stones[i - 1]];
// }
// }
// }
// // ans
// int minVal = Integer.MAX_VALUE;
// for (int j = 0; j <= sum; j++) {
// if (dp[n][j] && minVal > Math.abs(2 * j - sum)) {
// minVal = Math.abs(2 * j - sum);
// }
// }
// return minVal;
// }
// DP (Subset && Optimized to use 1D array)
public int lastStoneWeightII(int[] stones) {
int n = stones.length;
int sum = 0;
for (int i : stones) sum += i;
// subproblem: the sum of subset stones [0,j] with adding +,-, in front of each element, calculated sum is j
boolean[] dp = new boolean[sum + 1];
// init
dp[0] = true;
// recurrence relation
for (int i = 1; i <= n; i++) {
boolean[] temp = new boolean[sum + 1];
for (int j = 0; j <= sum; j++) {
temp[j] = dp[j];
if (j - stones[i - 1] >= 0) {
temp[j] |= dp[j - stones[i - 1]];
}
}
dp = temp;
}
// ans
int minVal = Integer.MAX_VALUE;
for (int j = 0; j <= sum; j++) {
if (dp[j] && minVal > Math.abs(2 * j - sum)) {
minVal = Math.abs(2 * j - sum);
}
}
return minVal;
}
}