Sunday, October 9, 2016

LeetCode 426 - Partition Equal Subset Sum


https://leetcode.com/problems/partition-equal-subset-sum/
Given a non-empty array containing only positive integers, find if the array can be partitioned into two subsets such that the sum of elements in both subsets is equal.
Note:
Both the array size and each of the array element will not exceed 100.
Example 1:
Input: [1, 5, 11, 5]

Output: true

Explanation: The array can be partitioned as [1, 5, 5] and [11].
Example 2:
Input: [1, 2, 3, 5]

Output: false

Explanation: The array cannot be partitioned into equal sum subsets.

X. DP
http://www.cnblogs.com/grandyang/p/5951422.html
这道题给了我们一个数组,问我们这个数组能不能分成两个非空子集合,使得两个子集合的元素之和相同。那么我们想,原数组所有数字和一定是偶数,不然根本无法拆成两个和相同的子集合,那么我们只需要算出原数组的数字之和,然后除以2,就是我们的target,那么问题就转换为能不能找到一个非空子集合,使得其数字之和为target。开始我想的是遍历所有子集合,算和,但是这种方法无法通过OJ的大数据集合。于是乎,动态规划DP就是我们的不二之选。我们定义一个一维的dp数组,其中dp[i]表示数字i是否是原数组的任意个子集合之和,那么我们我们最后只需要返回dp[target]就行了。我们初始化dp[0]为true,由于题目中限制了所有数字为正数,那么我们就不用担心会出现和为0或者负数的情况。那么关键问题就是要找出递归公式了,我们需要遍历原数组中的数字,对于遍历到的每个数字nums[i],我们需要更新我们的dp数组,要更新[nums[i], target]之间的值,那么对于这个区间中的任意一个数字j,如果dp[j - nums[j]]为true的话,那么dp[j]就一定为true,于是地推公式如下:
dp[j] = dp[j] || dp[j - nums[i]]         (nums[i] <= j <= target)

https://discuss.leetcode.com/topic/62307/java-short-dp-solution-with-explanation
If sum of all the numbers is odd, the surely we cannot reach equal partition.
Using a boolean dp array (limit its max index to sum/2) whose ith entry indicates there is a way to reach the value i using certain subset of the numbers. SO if at any time we can find a subset to reach sum/2 index, we are able to equally partition.
    public boolean canPartition(int[] nums) {
        // check edge case
        if (nums == null || nums.length == 0) {
            return true;
        }
        // preprocess
        int volumn = 0;
        for (int num : nums) {
            volumn += num;
        }
        if (volumn % 2 != 0) {
            return false;
        }
        volumn /= 2;
        // dp def
        boolean[] dp = new boolean[volumn + 1];
        // dp init
        dp[0] = true;
        // dp transition
        for (int i = 1; i <= nums.length; i++) { // if(nums[i-1] > volumn/2) retu
            for (int j = volumn; j >= nums[i-1]; j--) {
                dp[j] = dp[j] || dp[j - nums[i-1]];
            }
        }
        return dp[volumn];
    }
https://discuss.leetcode.com/topic/62913/38ms-dp-java-solution
    public boolean canPartition(int[] nums) {
        int n = nums.length;
        if (n == 0) 
            return true;
        int sum = 0;
        for (int num: nums) {
            sum += num;
        }
        if (sum % 2 == 1)
            return false;
        Arrays.sort(nums);
        int target = sum / 2;
        boolean[][] dp = new boolean[n + 1][target + 1];
        dp[0][0] = true;
        for (int i = 1; i <= n; i++) {
            if (nums[i-1] == target)
                return true;
            if (nums[i-1] > target)
                return false;
            System.arraycopy(dp[i-1], 0, dp[i], 0, Math.min(target + 1, nums[i-1]));
            for (int j = nums[i-1]; j <= target; j++) {
                dp[i][j] = dp[i-1][j - nums[i-1]];
            }
            if (dp[i][target])
                return true;
        }
        return false;
    }
https://discuss.leetcode.com/topic/64499/java-solution-similar-to-subset-sum-problem
It's similar to Subset Sum Problemwhich asks us to find if there is a subset whose sum equals to target value. For this problem, the target value is exactly the half of sum of array.
    public boolean canPartition(int[] nums) {
        int sum = 0;
        for(int num: nums) sum += num;
        if(sum % 2 == 1) return false;
        
        int target = sum / 2;
        boolean[][] dp = new boolean[nums.length][target + 1];
        // deal with the first row
        if(nums[0] <= target) dp[0][nums[0]] = true;
        
        // deal with the first col
        for(int i = 0; i < nums.length; i++) dp[i][0] = true;
        
        // deal with the rest
        for(int i = 1; i < dp.length; i++) {
            for(int j = 1; j < dp[0].length; j++) {
                if(j < nums[i]) {
                    dp[i][j] = dp[i - 1][j];
                } else {
                    dp[i][j] = dp[i - 1][j] || dp[i - 1][j - nums[i]];
                }
            }
        }
        return dp[dp.length - 1][dp[0].length - 1];
    }

http://bookshadow.com/weblog/2016/10/09/leetcode-partition-equal-subset-sum/
动态规划(Dynamic Programming)
利用数组dp[i]记录和为i的子数组是否存在,初始令dp[0] = 1
for num in nums:
    for i in range(sum(nums) - num + 1):
        if dp[i]: dp[i + num] = 1
def canPartition(self, nums): """ :type nums: List[int] :rtype: bool """ sums = sum(nums) if sums & 1: return False nset = set([0]) for n in nums: for m in nset.copy(): nset.add(m + n) return sums / 2 in nset

X. DP2
http://www.cnblogs.com/wangxiaobao/p/5943978.html
用01背包的思路,bool值 dp[i][j] 表示从0,1,2...i选重量为j是否可能。
则有递推关系:
dp[i][j] = dp[i - 1][j] || dp[i][j - nums[i]]; (i位置取或者不取)
结果返回值为dp[nums.size() - 1][sum]; (sum为和的一半)
 3     bool canPartition(vector<int>& nums) {
 4         int sum = 0;
 5         for (int i = 0; i < nums.size(); ++i) {
 6             sum += nums[i];
 7         }
 8         if (sum % 2 == 1) {
 9             return false;
10         } 
11         sum /= 2;
12         int dp[nums.size()][sum] = {false};
13         if (nums.size() == 1) {
14             return false;
15         }
16         for (int i = 0; i <= sum; ++i) {
17             if (nums[0] == i) {
18                 dp[0][i] = true;
19             }
20         }
21         for (int i = 1; i < nums.size(); ++i) {
22             for (int j = 1; j <= sum; ++j) {
23                 dp[i][j] = dp[i - 1][j] || dp[i][j - nums[i]];
24             }
25         }
26         return dp[nums.size() - 1][sum];
27     }

X. DFS+Cache - TLE
https://discuss.leetcode.com/topic/62267/java-solution-with-commets-using-dfs
    public boolean canPartition(int[] nums) {
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        int sum = 0;
        for(int i : nums){
            if(map.containsKey(i)){
                map.put(i, map.get(i) + 1);
            }else{
                map.put(i, 1);
            }
            sum += i;
        }
        if(sum % 2 == 1) return false;
        return helper(map, sum / 2);
    }
    
    private boolean helper(Map<Integer, Integer> map, int target){
        /*target is achieveable*/
        if(map.containsKey(target) && map.get(target) > 0) return true;
        /*dfs*/
        for(int key : map.keySet()){
            if(key < target && map.get(key) > 0){
                map.put(key, map.get(key) - 1);
                if(helper(map, target - key)) return true;
                map.put(key, map.get(key) + 1);
            }
        }
        return false;
    }
}
https://discuss.leetcode.com/topic/62342/why-dp-is-slower-than-dfs-even-with-no-memoization/
The N in DFS is the length of the input array, while in DP it is the half of the summation of all elements in the input array. It might be possible that the length of the input array is short, however, the summation is pretty large, in which case DP performs worse than DFS.
Could be. But then again: both size and elements have the same bound in the problem description (100). That means the maximum possible sum is O(N^2). That makes DP O(N^3), and backtracking is still O(2^N). Even when N is around 20 they should be on about the same level, for larger N backtracking should be prohibitively slow! And there are test cases as large as 50 elements.

X. DFS - Brute Force
http://blog.csdn.net/mebiuw/article/details/52765840
老旧的暴力法 仅供参考 不可AC
/** * 首先和为奇数的过滤 * 其次使用DFS * 排序后可以剪枝很多情况 * */ public boolean canPartition(int[] nums) { Arrays.sort(nums); int sum=0; for (int num:nums) sum+= num; if(sum % 2 == 1) return false; sum/=2; return dfs(0,sum,nums); } // 一一尝试 public boolean dfs(int index,int sum,int[] nums){ sum -= nums[index] ; if(sum == 0) return true; for(int i=index+1;i<nums.length;i++){ if(sum<nums[i]) break; if(dfs(i,sum,nums)) return true; } return false; }

X. BitSet
https://discuss.leetcode.com/topic/62334/simple-c-4-line-solution-using-a-bitset
http://www.cnblogs.com/grandyang/p/5951422.html
这道题还可以用bitset来做,感觉也十分的巧妙,bisets的大小设为5001,为啥呢,因为题目中说了数组的长度和每个数字的大小都不会超过100,那么最大的和为10000,那么一半就是5000,前面再加上个0,就是5001了。我们初始化把最低位赋值为1,然后我们算出数组之和,然后我们遍历数字,对于遍历到的数字num,我们把bits向左平移num位,然后再或上原来的bits,这样所有的可能出现的和位置上都为1。举个例子来说吧,比如对于数组[2,3]来说,初始化bits为1,然后对于数字2,bits变为101,我们可以看出来bits[2]标记为了1,然后遍历到3,bits变为了101101,我们看到bits[5],bits[3],bits[2]都分别为1了,正好代表了可能的和2,3,5,这样我们遍历玩整个数组后,去看bits[sum >> 1]是否为1即可
    bool canPartition(vector<int>& nums) {
        bitset<5001> bits(1);
        int sum = 0;
        for (auto n : nums) {
            sum += n;
            bits |= bits << n;
        }
        return !(sum & 1) && bits[sum >> 1]; // !(sum % 2) && bits[sum / 2];
    }


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