Submission for week 2

week02/Implement_strStr.cpp

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+//Problem 5: Implement strStr()
+
+//Solution:
+
+//Time Complexity: O(N) where N is the length of haystack string.
+//Space Complexity: O(1)
+
+int strStr(string haystack, string needle){
+    if(needle == "") return 0;
+    else {
+        int haystackLength = haystack.length();
+        int needleLength = needle.length();
+        for(int i = 0; i <= (haystackLength - needleLength); i++){
+            if(haystack[i] == needle[0] &&
+                haystack.substr(i, needleLength) == needle){
+                return i;
+            }
+        }
+        return -1;
+    }
+}

week02/Implement_strStr.md

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+#### Problem 5: Implement strStr()
+___
+
+##### Solution
+
+```c++
+//Time Complexity: O(N) where N is the length of haystack string.
+//Space Complexity: O(1)
+
+int strStr(string haystack, string needle){
+    if(needle == "") return 0;
+    else {
+        int haystackLength = haystack.length();
+        int needleLength = needle.length();
+        for(int i = 0; i <= (haystackLength - needleLength); i++){
+            if(haystack[i] == needle[0] &&
+                haystack.substr(i, needleLength) == needle){
+                return i;
+            }
+        }
+        return -1;
+    }
+}
+```

week02/Implement_strStr.txt

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+//Problem 5: Implement strStr()
+
+//Solution:
+
+//Time Complexity: O(N) where N is the length of haystack string.
+//Space Complexity: O(1)
+
+int strStr(string haystack, string needle){
+    if(needle == "") return 0;
+    else {
+        int haystackLength = haystack.length();
+        int needleLength = needle.length();
+        for(int i = 0; i <= (haystackLength - needleLength); i++){
+            if(haystack[i] == needle[0] &&
+                haystack.substr(i, needleLength) == needle){
+                return i;
+            }
+        }
+        return -1;
+    }
+}

week02/Longest_Common_Prefix.cpp

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+//Problem 3: Longest Common Prefix (Easy)
+
+//Solution:
+
+// Time complexity: O(n * m)
+// where, n = number of strings in the vector and
+// m = number of common characters ( length of common prefix )
+// Space complexity: O(1)
+
+string longestCommonPrefix(vector<string>& strs){
+    if(strs.size()) == 1){
+        return strs[0];
+    }
+    else{
+        int numberOfStrings = strs.size();
+        string currentLCP = strs[0];
+        for(int i = 1; i < numberOfStrings; i++){
+            int j;
+            for(int j = 0; j < strs[i].length(); j++){
+                if(strs[i][j] != currentLCP[j]) break;
+            }
+            currentLCP = currentLCP.substr(0, j);
+        }
+        return currentLCP;
+    }
+}

week02/Longest_Common_Prefix.md

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+#### Problem 3: Longest Common Prefix (Easy)
+___
+
+###### Solution
+
+```c++
+// Time complexity: O(n * m)
+// where, n = number of strings in the vector and
+// m = number of common characters ( length of common prefix )
+// Space complexity: O(1)
+
+string longestCommonPrefix(vector<string>& strs){
+    if(strs.size()) == 1){
+        return strs[0];
+    }
+    else{
+        int numberOfStrings = strs.size();
+        string currentLCP = strs[0];
+        for(int i = 1; i < numberOfStrings; i++){
+            int j;
+            for(int j = 0; j < strs[i].length(); j++){
+                if(strs[i][j] != currentLCP[j]) break;
+            }
+            currentLCP = currentLCP.substr(0, j);
+        }
+        return currentLCP;
+    }
+}
+```

week02/Longest_Common_Prefix.txt

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+//Problem 3: Longest Common Prefix (Easy)
+
+//Solution:
+
+// Time complexity: O(n * m)
+// where, n = number of strings in the vector and
+// m = number of common characters ( length of common prefix )
+// Space complexity: O(1)
+
+string longestCommonPrefix(vector<string>& strs){
+    if(strs.size()) == 1){
+        return strs[0];
+    }
+    else{
+        int numberOfStrings = strs.size();
+        string currentLCP = strs[0];
+        for(int i = 1; i < numberOfStrings; i++){
+            int j;
+            for(int j = 0; j < strs[i].length(); j++){
+                if(strs[i][j] != currentLCP[j]) break;
+            }
+            currentLCP = currentLCP.substr(0, j);
+        }
+        return currentLCP;
+    }
+}

week02/Longest_substring_length_without_repeating_characters.cpp

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+//Problem 1: Longest substring without repeating characters (Medium)
+
+//Solution:
+
+// Time complexity: O(n)
+// where, n = number of characters in the string and
+// Space complexity: O(1)
+
+int lengthOfLongestSubstring(string s){
+    vector<int> characterCount(256, 0);
+    int longestSubStrLen = 0;
+    int left = 0;
+    int right = 0;
+    int duplicateCount = 0;
+    int strLen = s.length();
+    while(right < strLen){
+        characterCount[s[right]]++;
+        if(characterCount[s[right]] > 1){
+            duplicateCount++;
+        }
+        right++;
+        while(duplicateCount > 0){
+            characterCount[s[left]]--;
+            if(characterCount[s[left]] == 1){
+                duplicateCount--;
+            }
+            left++;
+        }
+        longestSubStrLen = max(longestSubStrLen, right - left);
+    }
+    return longestSubStrLen;
+}

week02/Longest_substring_length_without_repeating_characters.md

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+#### Problem 1: Longest substring without repeating characters (Medium)
+___
+
+###### Solution
+
+```c++
+// Time complexity: O(n)
+// where, n = number of characters in the string and
+// Space complexity: O(1)
+
+int lengthOfLongestSubstring(string s){
+    vector<int> characterCount(256, 0);
+    int longestSubStrLen = 0;
+    int left = 0;
+    int right = 0;
+    int duplicateCount = 0;
+    int strLen = s.length();
+    while(right < strLen){
+        characterCount[s[right]]++;
+        if(characterCount[s[right]] > 1){
+            duplicateCount++;
+        }
+        right++;
+        while(duplicateCount > 0){
+            characterCount[s[left]]--;
+            if(characterCount[s[left]] == 1){
+                duplicateCount--;
+            }
+            left++;
+        }
+        longestSubStrLen = max(longestSubStrLen, right - left);
+    }
+    return longestSubStrLen;
+}
+```

week02/Longest_substring_length_without_repeating_characters.txt

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+//Problem 1: Longest substring without repeating characters (Medium)
+
+//Solution:
+
+// Time complexity: O(n)
+// where, n = number of characters in the string and
+// Space complexity: O(1)
+
+int lengthOfLongestSubstring(string s){
+    vector<int> characterCount(256, 0);
+    int longestSubStrLen = 0;
+    int left = 0;
+    int right = 0;
+    int duplicateCount = 0;
+    int strLen = s.length();
+    while(right < strLen){
+        characterCount[s[right]]++;
+        if(characterCount[s[right]] > 1){
+            duplicateCount++;
+        }
+        right++;
+        while(duplicateCount > 0){
+            characterCount[s[left]]--;
+            if(characterCount[s[left]] == 1){
+                duplicateCount--;
+            }
+            left++;
+        }
+        longestSubStrLen = max(longestSubStrLen, right - left);
+    }
+    return longestSubStrLen;
+}

week02/decode_string.cpp

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+//Problem 12: Decode String (Difficulty: Medium)
+
+//Solution:
+
+// Time Complexity: O(N)
+// Space Complexity: O(N)
+
+string decodingString(string& s, int& i){
+    string res;
+    int strLen = s.length();
+    while(i < strLen && s[i] != ']'){
+        // checking if s[i] is a digit
+        if(!isdigit(s[i]){
+            res += s[i];
+            i++;
+        }
+        else{
+            int num = 0;
+            while(i < strLen && isdigit(s[i])){
+                num = num * 10 + (s[i] - '0');
+                i++;
+            }
+            i++; // skipping '['
+            string tmp = decodingString(s, i);
+            i++; // skipping ']'
+
+            while(num > 0){
+                res += tmp;
+                num--;
+            }
+        }
+    }
+    return res;
+}
+
+string decodeString(string s){
+    int i = 0;
+    return decodingString(s, i);
+}

week02/decode_string.md

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+##### Problem 12: Decode String (Difficulty: Medium)
+___
+
+#### Solution
+
+```c++
+// Time Complexity: O(N)
+// Space Complexity: O(N)
+
+string decodingString(string& s, int& i){
+    string res;
+    int strLen = s.length();
+    while(i < strLen && s[i] != ']'){
+        // checking if s[i] is a digit
+        if(!isdigit(s[i]){
+            res += s[i];
+            i++;
+        }
+        else{
+            int num = 0;
+            while(i < strLen && isdigit(s[i])){
+                num = num * 10 + (s[i] - '0');
+                i++;
+            }
+            i++; // skipping '['
+            string tmp = decodingString(s, i);
+            i++; // skipping ']'
+
+            while(num > 0){
+                res += tmp;
+                num--;
+            }
+        }
+    }
+    return res;
+}
+
+string decodeString(string s){
+    int i = 0;
+    return decodingString(s, i);
+}
+```

week02/decode_string.txt

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+//Problem 12: Decode String (Difficulty: Medium)
+
+//Solution:
+
+// Time Complexity: O(N)
+// Space Complexity: O(N)
+
+string decodingString(string& s, int& i){
+    string res;
+    int strLen = s.length();
+    while(i < strLen && s[i] != ']'){
+        // checking if s[i] is a digit
+        if(!isdigit(s[i]){
+            res += s[i];
+            i++;
+        }
+        else{
+            int num = 0;
+            while(i < strLen && isdigit(s[i])){
+                num = num * 10 + (s[i] - '0');
+                i++;
+            }
+            i++; // skipping '['
+            string tmp = decodingString(s, i);
+            i++; // skipping ']'
+
+            while(num > 0){
+                res += tmp;
+                num--;
+            }
+        }
+    }
+    return res;
+}
+
+string decodeString(string s){
+    int i = 0;
+    return decodingString(s, i);
+}