## Problem Statement

Given a string **S**. The task is to find the lexicographically smallest string possible by inserting a given character.

A string **a** is lexicographically smaller than string **b** (of the same length) if in the first position where **a** and **b** differ, string a has a letter that appears earlier in the alphabet than the corresponding letter in b. For example, “abc” is lexicographically smaller than “bac” because ‘a’ comes before ‘b’ in dictionary order.

**Examples:**

**Input: **S = “axc”, ch = ‘b’**Output: **abxc**Explanation:**

Inserting character **b** in the second position gives the lexicographically smallest string.

**Input: ** S = “abcd”, ch = ‘e’**Output:** abcde**Explanation:**

Inserting character **e** in the last index gives the lexicographically smallest string.

## Approach

The approach is to insert the given character just before the position, where the **S[i]** is lexicographically larger than the character **ch**.

**Algorithm:**

- Iterate over the string from
**i = 0**till**i = N.** - If the current character of the string
**S[i]**is lexicographically larger than the character**ch,**insert the character at that index and terminate the loop. - Else, continue iterating and if no such character is found which is lexicographically greater than
**ch**, insert**ch**at the end of the string.

### C++ Implementation

### Java Implementation

### Python Implementation

**Time Complexity:** O(N), where N is the length of the string.**Space Complexity:** O(1), as no extra space is used.

## Practice Question

Lexicographically Largest Array

## Frequently Asked Questions

**What is the lexicographically smallest string?**

A string a is lexicographically smaller than string b (of the same length) if in the first position where a and b differ, string a has a letter that appears earlier in the alphabet than the corresponding letter in b

**What is the time complexity of the insertion of a character at a given position?**

The time complexity of insertion of a character at a given position is O(1).