Given an array of strings. The array has both empty and non-empty strings. All non-empty strings are in sorted order. Empty strings can be present anywhere between non-empty strings.

Examples:

Input : arr[] = {"for", "", "", "", "geeks", "ide", "", "practice", "" , "", "quiz", "", ""}; str = "quiz" Output : 10 The string "quiz" is present at index 10 in given array.

A **simple solution** is to linearly search given str in array of strings.

A **better solution** is to do modified Binary Search. Like normal binary search, we compare given str with middle string. If middle string is empty, we find the closest non-empty string x (by linearly searching on both sides). Once we find x, we do standard binary search, i.e., we compare given str with x. If str is same as x, we return index of x. if str is greater, we recur for right half, else we recur for left half.

Below is C++ implementation of the idea.

`// C++ program to find location of a str in ` `// an array of strings which is sorted and ` `// has empty strings between strings. ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Compare two string equals are not ` `int` `compareStrings(string str1, string str2) ` `{ ` ` ` `int` `i = 0; ` ` ` `while` `(str1[i] == str2[i] && str1[i] != ` `'\0'` `) ` ` ` `i++; ` ` ` `if` `(str1[i] > str2[i]) ` ` ` `return` `-1; ` ` ` ` ` `return` `(str1[i] < str2[i]); ` `} ` ` ` `// Main function to find string location ` `int` `searchStr(string arr[], string str, ` `int` `first, ` ` ` `int` `last) ` `{ ` ` ` `if` `(first > last) ` ` ` `return` `-1; ` ` ` ` ` `// Move mid to the middle ` ` ` `int` `mid = (last+first)/2; ` ` ` ` ` `// If mid is empty , find closet non-empty string ` ` ` `if` `(arr[mid].empty()) ` ` ` `{ ` ` ` `// If mid is empty, search in both sides of mid ` ` ` `// and find the closest non-empty string, and ` ` ` `// set mid accordingly. ` ` ` `int` `left = mid - 1; ` ` ` `int` `right = mid + 1; ` ` ` `while` `(` `true` `) ` ` ` `{ ` ` ` `if` `(left < first && right > last) ` ` ` `return` `-1; ` ` ` `if` `(right<=last && !arr[right].empty()) ` ` ` `{ ` ` ` `mid = right; ` ` ` `break` `; ` ` ` `} ` ` ` `if` `(left>=first && !arr[left].empty()) ` ` ` `{ ` ` ` `mid = left; ` ` ` `break` `; ` ` ` `} ` ` ` `right++; ` ` ` `left--; ` ` ` `} ` ` ` `} ` ` ` ` ` `// If str is found at mid ` ` ` `if` `(compareStrings(str, arr[mid]) == 0) ` ` ` `return` `mid; ` ` ` ` ` `// If str is greater than mid ` ` ` `if` `(compareStrings(str, arr[mid]) < 0) ` ` ` `return` `searchStr(arr, str, mid+1, last); ` ` ` ` ` `// If str is smaller than mid ` ` ` `return` `searchStr(arr, str, first, mid-1); ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `// Input arr of Strings. ` ` ` `string arr[] = {` `"for"` `, ` `""` `, ` `""` `, ` `""` `, ` `"geeks"` `, ` `"ide"` `, ` `""` `, ` ` ` `"practice"` `, ` `""` `, ` `""` `, ` `"quiz"` `, ` `""` `, ` `""` `}; ` ` ` ` ` `// input Search String ` ` ` `string str = ` `"quiz"` `; ` ` ` `int` `n = ` `sizeof` `(arr)/` `sizeof` `(arr[0]); ` ` ` ` ` `cout << searchStr(arr, str, 0, n-1); ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

Output:

10

Although this approach works better than linear search, the worst-case runtime for this algorithm is O(n).

This article is contributed by **Mr. Somesh Awasthi**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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