# Questions from 03/11/19 - 03/17/19

## Minimum Window Substring

76. Minimum Window Substring

### description

degree of difficulty: $\color{#e91e63}{hard}$

Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).

Example:

Input: S = “ADOBECODEBANC”, T = “ABC” Output: “BANC” Note:

• If there is no such window in S that covers all characters in T, return the empty string “”.
• If there is such window, you are guaranteed that there will always be only one unique minimum window in S.

### solution

Solution is from @zjh08177

It is hash table and two point question, and @zjh08177 gives us a template of this kind of questions

Here comes the template.

For most substring problem, we are given a string and need to find a substring of it which satisfy some restrictions. A general way is to use a hashmap assisted with two pointers. The template is given below.

int findSubstring(string s){
vector<int> map(128,0);
int counter; // check whether the substring is valid
int begin=0, end=0; //two pointers, one point to tail and one  head
int d; //the length of substring

for() { /* initialize the hash map here */ }

while(end<s.size()){

if(map[s[end++]]-- ?){  /* modify counter here */ }

while(/* counter condition */){

/* update d here if finding minimum*/

//increase begin to make it invalid/valid again

if(map[s[begin++]]++ ?){ /*modify counter here*/ }
}

/* update d here if finding maximum*/
}
return d;
}


So, come to this problem, we need to find the min-distance substring which contains all element of string T, so first of all, we need two points one represent the start point - begin, and one represent the end point - end, we need to find the substring whatever it is the min-distance one, and then, we need to look backwards, we move the begin point to make sure that we skip every uncorrelated element, and than begin comes to next step (like S = "DAOBECODEBANC", T = "ABC", begin first at index0, and we move it to index1, so we skipped the uncorrelated D, and next we move it to index2), so that we can start the next search, to make sure if there has shorter substring; for the end point, we only need to concerned that when we comes to the end point, the substring need to contains all T element, and every time, we change end first, and then change begin and calculate the min-distance (end - begin) in the meanwhile.

the cpp code:

class Solution {
public:
string minWindow(string s, string t) {
vector<int> map(128,0);
for(auto c: t) map[c]++;
int counter=t.size(), begin=0, end=0, d=INT_MAX, head=0;
while(end<s.size()){
if(map[s[end++]]-->0) counter--; //in t
while(counter==0){ //valid
if(map[s[begin++]]++==0) counter++;  //make it invalid
}
}
}
};


## Minimum Window Substring

42. Trapping Rain Water

### description

degree of difficulty: $\color{#e91e63}{hard}$

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!

Example:

Input: [0,1,0,2,1,0,1,3,2,1,2,1] Output: 6

### solution

Solution is from @mcrystal

It is a typical two point question. Usually, we would set two point - slow point and fast point start from beginning, but for this question, it will become a lit bit complex, and if we start from the left and right, we can have a easy-understanding solution.

We see the wall like a pool, we reduce this pool’s scale using a left point move from left to right, and a right point move from right to left; and in this process, we can get a new, higher wall that can spilt the pool into a new, smaller scale, or we just pass a lower stage and this lower stage can storage the water.

class Solution {
public:
int trap(vector<int>& height) {
int n =height.size();
int left=0; int right=n-1;
int res=0;
int maxleft=0, maxright=0;
while(left<=right){
if(height[left]<=height[right]){
if(height[left]>=maxleft) maxleft=height[left];
else res+=maxleft-height[left];
left++;
}
else{
if(height[right]>=maxright) maxright= height[right];
else res+=maxright-height[right];
right--;
}
}
return res;
}
};


## Edit Distance

72. Edit Distance

### description

degree of difficulty: $\color{#e91e63}{hard}$

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.

You have the following 3 operations permitted on a word:

• Insert a character
• Delete a character
• Replace a character

Example 1:

Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation:
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e') Example 2:

Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation:
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')


### solution

Solution is from @jianchao-li

This is a DP problem, we have two strings ready to compare, so it is better if we could use a 2D vector(one vector is good, but 2D is better to understand), we define the state dp[i][j] to be the minimum number of operations to convert word1[0..i) to word2[0..j).

For the base case, that is, to convert a string to an empty string, the mininum number of operations (deletions) is just the length of the string. So we have dp[i][0] = i and dp[0][j] = j.

For the general case to convert word1[0..i) to word2[0..j), we break this problem down into sub-problems. Suppose we have already known how to convert word1[0..i - 1) to word2[0..j - 1) (dp[i - 1][j - 1]), if word1[i - 1] == word2[j - 1], then no more operation is needed and dp[i][j] = dp[i - 1][j - 1].

If word1[i - 1] != word2[j - 1], we need to consider three cases.

Replace word1[i - 1] by word2[j - 1] (dp[i][j] = dp[i - 1][j - 1] + 1);

If word1[0..i - 1) = word2[0..j), means that word1[0..i - 1) convert word2[0..j - 1), and the redundant is the word1[i - 1], so we then delete word1[i - 1] (dp[i][j] = dp[i - 1][j] + 1);

If word1[0..i) + word2[j - 1] = word2[0..j) then insert word2[j - 1] to word1[0..i) (dp[i][j] = dp[i][j - 1] + 1). So when word1[i - 1] != word2[j - 1], dp[i][j] will just be the minimum of the above three cases.

class Solution {
public:
int minDistance(string word1, string word2) {
int m = word1.size(), n = word2.size();
vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
for (int i = 1; i <= m; i++) {
dp[i][0] = i;
}
for (int j = 1; j <= n; j++) {
dp[0][j] = j;
}
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (word1[i - 1] == word2[j - 1]) {
dp[i][j] = dp[i - 1][j - 1];
} else {
dp[i][j] = min(dp[i - 1][j - 1], min(dp[i][j - 1], dp[i - 1][j])) + 1;
}
}
}
return dp[m][n];
}
};