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Minimum Window Substring in C++
Suppose we have a string S and T. We have to find the minimum window in S which will contain all the characters in T. So if the input is like “ABHDAXCVBAGTXATYCB” and T = “ABC”, then the result will be: “CVBA”.
To solve this, we will follow these steps −
Create one map m
store the frequency of x into m
length := size of s, left := 0, right := 0, ansLeft := 0 and ansRight := 0
counter := size of x, flag := false, ans := empty string
while height < size of s −
c := s[right]
if c is present in m, then
if m[c] > 0, then decrease counter by 1
decrease m[c] by 1
while counter = 0 and left <= right
if right – left + 1 <= length
length := right – left + 1
flag := true
ansLeft := left, ansRight := right
if left = right, then break the loop
c := s[left]
if c is present in m, then increase m[c] by 1
if m[c] > 0, then increase counter by 1
increase left by 1
increase right by 1
if flag is false, then return ans
otherwise for i in range ansLeft to ansRight, increase ans by s[i]
return ans
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
string minWindow(string s, string x) {
map <char, int> m;
for(int i =0;i<x.size();i++)m[x[i]]++;
int length = s.size();
int left = 0, right = 0 , ansLeft = 0, ansRight = 0;
int counter = x.size();
bool flag = false;
string ans = "";
while(right<s.size()){
char c = s[right];
if(m.find(c)!=m.end()){
if(m[c]>0)counter--;
m[c]--;
}
while(counter == 0 && left<=right){
if(right-left+1 <=length){
length = right-left+1;
flag = true;
ansLeft = left;
ansRight = right;
}
if(left == right)break;
c = s[left];
if(m.find(c)!=m.end()){
m[c]++;
if(m[c]>0)counter++;
}
left++;
}
right++;
}
if(!flag)return ans;
else
for(int i =ansLeft;i<=ansRight;i++)ans+=s[i];
return ans;
}
};
main(){
Solution ob;
cout << (ob.minWindow("ABHDAXCVBAGTXATYCB", "ABC"));
}Input
"ABHDAXCVBAGTXATYCB" "ABC"
Output
CVBA
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