Sunday, May 20, 2018

LeetCode 727 - Minimum Window Subsequence


http://www.cnblogs.com/lightwindy/p/8486724.html
Given strings S and T, find the minimum (contiguous) substring W of S, so that T is a subsequenceof W.
If there is no such window in S that covers all characters in T, return the empty string "". If there are multiple such minimum-length windows, return the one with the left-most starting index.
Example 1:
Input: 
S = "abcdebdde", T = "bde"
Output: "bcde"
Explanation: 
"bcde" is the answer because it occurs before "bdde" which has the same length.
"deb" is not a smaller window because the elements of T in the window must occur in order.
Note:
  • All the strings in the input will only contain lowercase letters.
  • The length of S will be in the range [1, 20000].
  • The length of T will be in the range [1, 100].


At time j, for each position e in S (e for end), let's remember the largest index cur[e] = s (for start) so that S[s: e+1] has T[:j] as a subsequence, and -1 (or None) otherwise if it isn't possible.
To update our knowledge as j += 1, if S[i] == T[j], then new[e] is last, the largest s we have seen so far (representing that T[:j] was found). We can prove this is just the most recent valid index we have seen.
At the end, we find the best answer: cur[e] = s means there was a window S[s: e+1]. In Python, we use cur and new, while in Java we use dp[j] and dp[~j] to keep track of the last two rows of our dynamic programming.
  1.     public String minWindow(String S, String T) {  
  2.         int[][] dp = new int[2][S.length()];  
  3.   
  4.         for (int i = 0; i < S.length(); ++i)  
  5.             dp[0][i] = S.charAt(i) == T.charAt(0) ? i : -1;  
  6.   
  7.         for (int j = 1; j < T.length(); ++j) {  
  8.             int last = -1;  
  9.             Arrays.fill(dp[j & 1], -1);  
  10.             for (int i = 0; i < S.length(); ++i) {  
  11.                 if (last >= 0 && S.charAt(i) == T.charAt(j))  
  12.                     dp[j & 1][i] = last;  
  13.                 if (dp[~j & 1][i] >= 0)  
  14.                     last = dp[~j & 1][i];  
  15.             }  
  16.         }  
  17.   
  18.         int start = 0, end = S.length();  
  19.         for (int e = 0; e < S.length(); ++e) {  
  20.             int s = dp[~T.length() & 1][e];  
  21.             if (s >= 0 && e - s < end - start) {  
  22.                 start = s;  
  23.                 end = e;  
  24.             }  
  25.         }  
  26.         return end < S.length() ? S.substring(start, end+1) : "";  
  27.     }  
https://www.jianshu.com/p/beaa7079a0fb
用dynamic programing来做,dp[i][j]表示T[0, j]是S[0,i]的subsequence, 所以我们的目标函数就是min(i-dp[i][n-1]) for all i < m
    public String minWindow(String S, String T) {
        int m = T.length(), n = S.length();
        int[][] dp = new int[m + 1][n + 1];
        for (int j = 0; j <= n; j++) {
            dp[0][j] = j + 1;
        }
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (T.charAt(i - 1) == S.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    dp[i][j] = dp[i][j - 1];
            }
        }
        }

        int start = 0, len = n + 1;
        for (int j = 1; j <= n; j++) {
            if (dp[m][j] != 0) {
                if (j - dp[m][j] + 1 < len) {
                    start = dp[m][j] - 1;
                    len = j - dp[m][j] + 1;
                }
            }
        }
        return len == n + 1 ? "" : S.substring(start, start + len);
    }

http://www.cnblogs.com/lightwindy/p/8486724.html
给定字符串S和T,在S中寻找最小连续子串W,使得T是W的子序列。如果没有找到返回"",如果找到多个最小长度的子串,返回左 index 最小的。
解法1:暴力搜索brute force,对于每一个s[i],从s[0]到s[i]扫描,看是否按顺序满足目标字符。 显然要超时,不是题目要求的。
解法2: 动态规划DP,  二维数组dp[i][j]表示T[0...i]在S中找到的起始下标index,使得S[index, j]满足目前T[0...i]。首先找到能满足满足T中第一个字符T[0]的S中的字符下标存入dp[0][j],也就是满足第一个字符要求一定是从这些找到的字符开始的。然后在开始找第二个字符T[1],扫到的字符dp[j]存有index,说明可以从这里记录的index开始,找到等于T[1]的S[j]就把之前那个index存进来,说明从这个index到j满足T[0..1],一直循环,直到T中的i个字符找完。如果此时dp[i][j]中有index,说明S[index, j]满足条件,如有多个输出最先找到的。
State: dp[i][j],表示在S中找到的起始下标 index ,使得 S[index...j] 满足目前 T[0...i] 是其子序列。
function: dp[i+1][k] = dp[i][j]  if S[k] = T[i+1] , 如果查看到第i+1行(也就是第 T[i+1]  的字符),如果满足S[k] = T[i+1],就把上一行找到的index赋给它。
Initialize: dp[0][j] = j if S[j] = T[0] , 二维数组的第一行,如果字符S[j] = T[0], 就把S[j]的index(就是j)付给它。其他元素均为 None 或者 -1。
Return:  dp[len(T) - 1][j], if  dp[len(T) - 1][j] != None, 返回最小的。如果没有返回 ""
由于我们只用到前一行的值,所以可以只用2行的二维数组,每一个循环更新其中的一行。可以用 j % 2 来往复使用。
    public String minWindow(String S, String T) {
        int m = T.length(), n = S.length();
        int[][] dp = new int[m + 1][n + 1];
        for (int j = 0; j <= n; j++) {
            dp[0][j] = j + 1;
        }
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (T.charAt(i - 1) == S.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1];
                else {
                    dp[i][j] = dp[i][j - 1];
            }
        }
        }
 
        int start = 0, len = n + 1;
        for (int j = 1; j <= n; j++) {
            if (dp[m][j] != 0) {
                if (j - dp[m][j] + 1 < len) {
                    start = dp[m][j] - 1;
                    len = j - dp[m][j] + 1;
                }
            }
        }
        return len == n + 1 "" : S.substring(start, start + len);
    }



Java:  DP, Time O(s*t), Space O(s*2)
    public String minWindow(String S, String T) { 
        int[][] dp = new int[2][S.length()]; 
   
        for (int i = 0; i < S.length(); ++i) 
            dp[0][i] = S.charAt(i) == T.charAt(0) ? i : -1
   
        for (int j = 1; j < T.length(); ++j) { 
            int last = -1
            Arrays.fill(dp[j & 1], -1); 
            for (int i = 0; i < S.length(); ++i) { 
                if (last >= 0 && S.charAt(i) == T.charAt(j)) 
                    dp[j & 1][i] = last; 
                if (dp[j & 1][i] >= 0
                    last = dp[j & 1][i]; 
            
        
   
        int start = 0, end = S.length(); 
        for (int e = 0; e < S.length(); ++e) { 
            int s = dp[T.length() & 1][e]; 
            if (s >= 0 && e - s < end - start) { 
                start = s; 
                end = e; 
            
        
        return end < S.length() ? S.substring(start, end+1) : ""
    


x.

A naive brute force is relatively easy: for each starting position S[i], scan left to right trying to match elements T[j] in order. Unfortunately, this is O(S^2)O(S2) complexity, so we seek to improve it.

Java: brute force, Time O(s*t), Space O(s*t)
    public String minWindow(String S, String T) { 
        int min = -1, idx = -1
        char[] Tc = T.toCharArray(); 
        char[] Sc = S.toCharArray(); 
        for(int i = 0;i < S.length();i++){ 
            if(Sc[i] != Tc[0]) continue
            int len = check(Tc,Sc,i); 
            if(len <= 0break
            if(min == -1 || len < min){ 
                idx = i; 
                min = len; 
            
        
        if(min == -1return ""
        return S.substring(idx, idx + min); 
    
       
    public int check(char[] Tc, char[] Sc, int start){ 
        int i = start, j = 0
        while(i < Sc.length && j < Tc.length){ 
            if(Sc[i] == Tc[j]) j++; 
            i++; 
        
        if(j == Tc.length) return i - start; 
         
        return -1
    

https://blog.csdn.net/zjucor/article/details/78511774
  1.     public String minWindow(String S, String T) {  
  2.         char[] cs = S.toCharArray();  
  3.         char[] ct = T.toCharArray();          
  4.         int lo = T.length(), hi = S.length();  
  5.         int min = Integer.MAX_VALUE, start = -1;  
  6.           
  7.         while(lo < hi) {  
  8.             int mid = (lo+hi)/2;  
  9.             int t = ok(cs, ct, mid);  
  10.             if(t != -1) {  
  11.                 min = mid;  
  12.                 start = t;  
  13.                 hi = mid;  
  14.             } else {  
  15.                 lo = mid+1;  
  16.             }  
  17.         }  
  18.           
  19.         return min==Integer.MAX_VALUE ? "" : S.substring(start, start+min);  
  20.     }  
  21.   
  22.     private int ok(char[] cs, char[] ct, int mid) {  
  23.         for(int i=0; i<cs.length-mid+1; i++) {  
  24.             if(isSeq(cs, i, i+mid-1, ct))  
  25.                 return i;  
  26.         }  
  27.         return -1;  
  28.     }  
  29.   
  30.     private boolean isSeq(char[] cs, int s, int t, char[] ct) {  
  31.         int i = s, j = 0;  
  32.         while(i <= t) {  
  33.             while(i<=t && cs[i]!=ct[j])  i++;  
  34.             if(i > t) break;  
  35.             i++;  
  36.             j++;  
  37.             if(j == ct.length)  return true;  
  38.         }  
  39.         return false;  
  40.     }  









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