Longest Common Prefix

first iteration

class Solution {public:    string longestCommonPrefix(vector<string> &strs) {                string prefix;                // empty string vector        if (strs.empty()) {            return prefix;        }                int idx=0;        char c;        while (true) {                        // get character to be compared from first string            if (idx < strs[0].size()) {                c = strs[0][idx];            } else {                break;            }                        // check if all strings has that character in the index            //   - break out if either out of length or not same            bool same = true;            for (int i = 1; i < strs.size(); i++) {                if (idx >= strs[i].size() || strs[i][idx] != c) {                    same = false;                    break;                }             }                        // grow prefix if it is same otherwise, break out            if (same) {                prefix.push_back(c);            } else {                break;            }                        idx++;        }                return prefix;    }};

2nd iteration with a little refactor

class Solution {public:    string longestCommonPrefix(vector<string> &strs) {                string prefix;                // empty string vector        if (strs.empty()) {            return prefix;        }                int len=0;        while (true) {            char var;            int i = 0;            for (; i < strs.size(); i++) {                // get character to be compared from first string                // could be '\0', but we break out as below                if (i==0) {                    var = strs[0][len];                }                                // break out if run out of length or character comparision                // failed                if (len == strs[i].size() || strs[i][len] != var) {                    break;                }             }                        // this means compare with this character failed, break out             // and return whatever we got for prefix            // NOTE: use the fact we early break out from for-loop or not to             // determine if we need continue the check            if (i != strs.size()) {                break;            }                        // continue the checkfor longest common prefix            prefix.push_back(var);            len++;        }                return prefix;    }};