Minimum Window Substring

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).

For example,
S = "ADOBECODEBANC"
T = "ABC"

Minimum window is "BANC".

Note:
If there is no such window in S that covers all characters in T, return the emtpy string "".

If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

class Solution {public:    int greater(int a[], int an, int b[], int bn){        int i;if(an < bn){return false;}        for(i = 0; i < 256; i++){            if(a[i] < b[i]){                return false;            }        }        return true;    }    string minWindow(string S, string T) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int tflag[256];        int sflag[256];        memset(tflag, 0, 256 * sizeof(int));        memset(sflag, 0, 256 * sizeof(int));        int i, j;        int lens = S.size();        int lent = T.size();        for(i =0; i < lent; i++){            tflag[T[i]]++;        }        vector<int> wind(lens, 0);        int exist = false;        for(i = 0; i < lens && !exist; i++){            sflag[S[i]]++;            if(greater(sflag, i+1, tflag, lent)){                exist = true;                wind[i] = i + 1;                break;            }        }        if(!exist){            return "";        }        for(j = 0; j <= i; j++){            sflag[S[j]]--;            if(greater(sflag, i - j + 1, tflag, lent)){                            } else {                wind[i] = i - j + 1;                sflag[S[j]]++;                break;            }        }                for(i++; i < lens; i++){            if(tflag[S[i]]){                if(S[i] == S[i - wind[i - 1]]){                    for(j = i - wind[i - 1] + 1; j < i; j++){                        if(tflag[S[j]]){                           if(sflag[S[j]] > tflag[S[j]]){                               sflag[S[j]]--;                           } else {                               break;                           }                        }                    }                    wind[i] = i - j + 1;                } else {                    sflag[S[i]]++;                    wind[i] = wind[i-1] + 1;                }            } else {                wind[i] = wind[i-1] + 1;            }        }        int minwind = 50000;        int pos=0;        for(i = 0; i < lens; i++){            if(wind[i] && wind[i] < minwind){                minwind= wind[i];                pos = i;            }        }        return S.substr(pos - wind[pos] + 1, wind[pos]);    }};