Given a string **str** consisting of **0, 1**, and** ***, the task is to find the maximum occurring character out of **0** and **1** after performing the given operations:

- Replace
*with0where*appears on the left side of the existing0s in the string.- Replace
*with1where*appears on the right side of the existing1s in the string.- If any
*can be replaced by both0and1, then it remains unchanged.

**Note: **If the frequency of 0 and 1 is same after performing the given operations then print **-1**.

**Examples:**

Input:str = “**0**1***0”Output:0Explanation:

Since0can replace the*to its left and1can replace the*to its right thus string becomes 000**1***0

Input:str = “0*1”Output:-1Explanation:

Both 0 and 1 have the same frequency hence the output is -1.

**Approach: **The idea to generate the final resultant string and then compare the frequency of 0 and 1. Below are the steps:

- Count the initial frequencies of 0 and 1 in the string and store them in variables say
**count_0**and**count_1**. - Initialize a variable, say
**prev**, as -1. Iterate over the string and check if the current character is*****. If so, then continue. - If it is the first character encountered and is
**0**then add all*****to**count_0**and change**prev**to current index. - Otherwise, if the first character is
**1**then change**prev**to current index. - If the previous character is
**1**and the current character is**0**then add half of*****in between the characters to**0**and half to**1**. - If the previous character is 0 and the current character is
**1**then no*****character in between them can be replaced. - If the previous and current both characters are of the same type then add the count of
*****to the frequencies. - Compare the frequencies of
**0**and**1**and print the maximum occurring character.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` ` ` `// Function to find the` `// maximum occurring character` `void` `solve(string S)` `{` ` ` `// Initialize count of` ` ` `// zero and one` ` ` `int` `count_0 = 0, count_1 = 0;` ` ` ` ` `int` `prev = -1;` ` ` ` ` `// Iterate over the given string` ` ` `for` `(` `int` `i = 0; i < S.length(); i++) {` ` ` ` ` `// Count the zeros` ` ` `if` `(S[i] == ` `'0'` `)` ` ` `count_0++;` ` ` ` ` `// Count the ones` ` ` `else` `if` `(S[i] == ` `'1'` `)` ` ` `count_1++;` ` ` `}` ` ` ` ` `// Iterate over the given string` ` ` `for` `(` `int` `i = 0; i < S.length(); i++) {` ` ` ` ` `// Check if character` ` ` `// is * then continue` ` ` `if` `(S[i] == ` `'*'` `)` ` ` `continue` `;` ` ` ` ` `// Check if first character` ` ` `// after * is X` ` ` `else` `if` `(S[i] == ` `'0'` `&& prev == -1) {` ` ` ` ` `// Add all * to` ` ` `// the frequency of X` ` ` `count_0 = count_0 + i;` ` ` ` ` `// Set prev to the` ` ` `// i-th character` ` ` `prev = i;` ` ` `}` ` ` ` ` `// Check if first character` ` ` `// after * is Y` ` ` `else` `if` `(S[i] == ` `'1'` `&& prev == -1) {` ` ` ` ` `// Set prev to the` ` ` `// i-th character` ` ` `prev = i;` ` ` `}` ` ` ` ` `// Check if prev character is 1` ` ` `// and current character is 0` ` ` `else` `if` `(S[prev] == ` `'1'` `&& S[i] == ` `'0'` `) {` ` ` ` ` `// Half of the * will be` ` ` `// converted to 0` ` ` `count_0 = count_0 + (i - prev - 1) / 2;` ` ` ` ` `// Half of the * will be` ` ` `// converted to 1` ` ` `count_1 = count_1 + (i - prev - 1) / 2;` ` ` `prev = i;` ` ` `}` ` ` ` ` `// Check if prev and current are 1` ` ` `else` `if` `(S[prev] == ` `'1'` `&& S[i] == ` `'1'` `) {` ` ` ` ` `// All * will get converted to 1` ` ` `count_1 = count_1 + (i - prev - 1);` ` ` `prev = i;` ` ` `}` ` ` ` ` `// No * can be replaced` ` ` `// by either 0 or 1` ` ` `else` `if` `(S[prev] == ` `'0'` `&& S[i] == ` `'1'` `)` ` ` ` ` `// Prev becomes the ith character` ` ` `prev = i;` ` ` ` ` `// Check if prev and current are 0` ` ` `else` `if` `(S[prev] == ` `'0'` `&& S[i] == ` `'0'` `) {` ` ` ` ` `// All * will get converted to 0` ` ` `count_0 = count_0 + (i - prev - 1);` ` ` `prev = i;` ` ` `}` ` ` `}` ` ` ` ` `// If frequency of 0` ` ` `// is more` ` ` `if` `(count_0 > count_1)` ` ` `cout << ` `"0"` `;` ` ` ` ` `// If frequency of 1` ` ` `// is more` ` ` `else` `if` `(count_1 > count_0)` ` ` `cout << ` `"1"` `;` ` ` ` ` `else` `{` ` ` `cout << -1;` ` ` `}` `}` ` ` `// Driver code` `int` `main()` `{` ` ` `// Given string` ` ` `string str = ` `"**0**1***0"` `;` ` ` ` ` `// Function Call` ` ` `solve(str);` ` ` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.io.*;` ` ` `class` `GFG{` ` ` `// Function to find the` `// maximum occurring character` `static` `void` `solve(String S)` `{` ` ` ` ` `// Initialize count of` ` ` `// zero and one` ` ` `int` `count_0 = ` `0` `, count_1 = ` `0` `;` ` ` ` ` `int` `prev = -` `1` `;` ` ` ` ` `// Iterate over the given string` ` ` `for` `(` `int` `i = ` `0` `; i < S.length(); i++)` ` ` `{` ` ` ` ` `// Count the zeros` ` ` `if` `(S.charAt(i) == ` `'0'` `)` ` ` `count_0++;` ` ` ` ` `// Count the ones` ` ` `else` `if` `(S.charAt(i) == ` `'1'` `)` ` ` `count_1++;` ` ` `}` ` ` ` ` `// Iterate over the given string` ` ` `for` `(` `int` `i = ` `0` `; i < S.length(); i++) ` ` ` `{` ` ` ` ` `// Check if character` ` ` `// is * then continue` ` ` `if` `(S.charAt(i) == ` `'*'` `)` ` ` `continue` `;` ` ` ` ` `// Check if first character` ` ` `// after * is X` ` ` `else` `if` `(S.charAt(i) == ` `'0'` `&& prev == -` `1` `)` ` ` `{` ` ` ` ` `// Add all * to` ` ` `// the frequency of X` ` ` `count_0 = count_0 + i;` ` ` ` ` `// Set prev to the` ` ` `// i-th character` ` ` `prev = i;` ` ` `}` ` ` ` ` `// Check if first character` ` ` `// after * is Y` ` ` `else` `if` `(S.charAt(i) == ` `'1'` `&& prev == -` `1` `)` ` ` `{` ` ` ` ` `// Set prev to the` ` ` `// i-th character` ` ` `prev = i;` ` ` `}` ` ` ` ` `// Check if prev character is 1` ` ` `// and current character is 0` ` ` `else` `if` `(S.charAt(prev) == ` `'1'` `&&` ` ` `S.charAt(i) == ` `'0'` `) ` ` ` `{` ` ` ` ` `// Half of the * will be` ` ` `// converted to 0` ` ` `count_0 = count_0 + (i - prev - ` `1` `) / ` `2` `;` ` ` ` ` `// Half of the * will be` ` ` `// converted to 1` ` ` `count_1 = count_1 + (i - prev - ` `1` `) / ` `2` `;` ` ` `prev = i;` ` ` `}` ` ` ` ` `// Check if prev and current are 1` ` ` `else` `if` `(S.charAt(prev) == ` `'1'` `&&` ` ` `S.charAt(i) == ` `'1'` `)` ` ` `{` ` ` ` ` `// All * will get converted to 1` ` ` `count_1 = count_1 + (i - prev - ` `1` `);` ` ` `prev = i;` ` ` `}` ` ` ` ` `// No * can be replaced` ` ` `// by either 0 or 1` ` ` `else` `if` `(S.charAt(prev) == ` `'0'` `&& ` ` ` `S.charAt(i) == ` `'1'` `)` ` ` ` ` `// Prev becomes the ith character` ` ` `prev = i;` ` ` ` ` `// Check if prev and current are 0` ` ` `else` `if` `(S.charAt(prev) == ` `'0'` `&& ` ` ` `S.charAt(i) == ` `'0'` `) ` ` ` `{` ` ` ` ` `// All * will get converted to 0` ` ` `count_0 = count_0 + (i - prev - ` `1` `);` ` ` `prev = i;` ` ` `}` ` ` `}` ` ` ` ` `// If frequency of 0` ` ` `// is more` ` ` `if` `(count_0 > count_1)` ` ` `System.out.print(` `"0"` `);` ` ` ` ` `// If frequency of 1` ` ` `// is more` ` ` `else` `if` `(count_1 > count_0)` ` ` `System.out.print(` `"1"` `);` ` ` ` ` `else` ` ` `{` ` ` `System.out.print(` `"-1"` `);` ` ` `}` `}` ` ` `// Driver code` `public` `static` `void` `main (String[] args)` `{` ` ` ` ` `// Given string` ` ` `String str = ` `"**0**1***0"` `;` ` ` ` ` `// Function call` ` ` `solve(str);` `}` `}` ` ` `// This code is contributed by code_hunt` |

## Python3

`# Python3 program for the above approach` ` ` `# Function to find the` `# maximum occurring character` `def` `solve(S):` ` ` ` ` `# Initialize count of` ` ` `# zero and one` ` ` `count_0 ` `=` `0` ` ` `count_1 ` `=` `0` ` ` ` ` `prev ` `=` `-` `1` ` ` ` ` `# Iterate over the given string` ` ` `for` `i ` `in` `range` `(` `len` `(S)) :` ` ` ` ` `# Count the zeros` ` ` `if` `(S[i] ` `=` `=` `'0'` `):` ` ` `count_0 ` `+` `=` `1` ` ` ` ` `# Count the ones` ` ` `elif` `(S[i] ` `=` `=` `'1'` `):` ` ` `count_1 ` `+` `=` `1` ` ` ` ` `# Iterate over the given string` ` ` `for` `i ` `in` `range` `(` `len` `(S)):` ` ` ` ` `# Check if character` ` ` `# is * then continue` ` ` `if` `(S[i] ` `=` `=` `'*'` `):` ` ` `continue` ` ` ` ` `# Check if first character` ` ` `# after * is X` ` ` `elif` `(S[i] ` `=` `=` `'0'` `and` `prev ` `=` `=` `-` `1` `):` ` ` ` ` `# Add all * to` ` ` `# the frequency of X` ` ` `count_0 ` `=` `count_0 ` `+` `i` ` ` ` ` `# Set prev to the` ` ` `# i-th character` ` ` `prev ` `=` `i` ` ` ` ` `# Check if first character` ` ` `# after * is Y` ` ` `elif` `(S[i] ` `=` `=` `'1'` `and` `prev ` `=` `=` `-` `1` `):` ` ` ` ` `# Set prev to the` ` ` `# i-th character` ` ` `prev ` `=` `i` ` ` ` ` `# Check if prev character is 1` ` ` `# and current character is 0` ` ` `elif` `(S[prev] ` `=` `=` `'1'` `and` `S[i] ` `=` `=` `'0'` `):` ` ` ` ` `# Half of the * will be` ` ` `# converted to 0` ` ` `count_0 ` `=` `count_0 ` `+` `(i ` `-` `prev ` `-` `1` `) ` `/` `2` ` ` ` ` `# Half of the * will be` ` ` `# converted to 1` ` ` `count_1 ` `=` `count_1 ` `+` `(i ` `-` `prev ` `-` `1` `) ` `/` `/` `2` ` ` `prev ` `=` `i` ` ` ` ` `# Check if prev and current are 1` ` ` `elif` `(S[prev] ` `=` `=` `'1'` `and` `S[i] ` `=` `=` `'1'` `):` ` ` ` ` `# All * will get converted to 1` ` ` `count_1 ` `=` `count_1 ` `+` `(i ` `-` `prev ` `-` `1` `)` ` ` `prev ` `=` `i` ` ` ` ` `# No * can be replaced` ` ` `# by either 0 or 1` ` ` `elif` `(S[prev] ` `=` `=` `'0'` `and` `S[i] ` `=` `=` `'1'` `):` ` ` ` ` `# Prev becomes the ith character` ` ` `prev ` `=` `i` ` ` ` ` `# Check if prev and current are 0` ` ` `elif` `(S[prev] ` `=` `=` `'0'` `and` `S[i] ` `=` `=` `'0'` `):` ` ` ` ` `# All * will get converted to 0` ` ` `count_0 ` `=` `count_0 ` `+` `(i ` `-` `prev ` `-` `1` `)` ` ` `prev ` `=` `i` ` ` ` ` `# If frequency of 0` ` ` `# is more` ` ` `if` `(count_0 > count_1):` ` ` `print` `(` `"0"` `)` ` ` ` ` `# If frequency of 1` ` ` `# is more` ` ` `elif` `(count_1 > count_0):` ` ` `print` `(` `"1"` `)` ` ` `else` `:` ` ` `print` `(` `"-1"` `)` ` ` `# Driver code` ` ` `# Given string` `str` `=` `"**0**1***0"` ` ` `# Function call` `solve(` `str` `)` ` ` `# This code is contributed by code_hunt` |

## C#

`// C# program for the above approach ` `using` `System;` ` ` `class` `GFG{` ` ` `// Function to find the` `// maximum occurring character` `static` `void` `solve(` `string` `S)` `{` ` ` ` ` `// Initialize count of` ` ` `// zero and one` ` ` `int` `count_0 = 0, count_1 = 0;` ` ` ` ` `int` `prev = -1;` ` ` ` ` `// Iterate over the given string` ` ` `for` `(` `int` `i = 0; i < S.Length; i++)` ` ` `{` ` ` ` ` `// Count the zeros` ` ` `if` `(S[i] == ` `'0'` `)` ` ` `count_0++;` ` ` ` ` `// Count the ones` ` ` `else` `if` `(S[i] == ` `'1'` `)` ` ` `count_1++;` ` ` `}` ` ` ` ` `// Iterate over the given string` ` ` `for` `(` `int` `i = 0; i < S.Length; i++)` ` ` `{` ` ` ` ` `// Check if character` ` ` `// is * then continue` ` ` `if` `(S[i] == ` `'*'` `)` ` ` `continue` `;` ` ` ` ` `// Check if first character` ` ` `// after * is X` ` ` `else` `if` `(S[i] == ` `'0'` `&& prev == -1) ` ` ` `{` ` ` ` ` `// Add all * to` ` ` `// the frequency of X` ` ` `count_0 = count_0 + i;` ` ` ` ` `// Set prev to the` ` ` `// i-th character` ` ` `prev = i;` ` ` `}` ` ` ` ` `// Check if first character` ` ` `// after * is Y` ` ` `else` `if` `(S[i] == ` `'1'` `&& prev == -1)` ` ` `{` ` ` ` ` `// Set prev to the` ` ` `// i-th character` ` ` `prev = i;` ` ` `}` ` ` ` ` `// Check if prev character is 1` ` ` `// and current character is 0` ` ` `else` `if` `(S[prev] == ` `'1'` `&& S[i] == ` `'0'` `)` ` ` `{` ` ` ` ` `// Half of the * will be` ` ` `// converted to 0` ` ` `count_0 = count_0 + (i - prev - 1) / 2;` ` ` ` ` `// Half of the * will be` ` ` `// converted to 1` ` ` `count_1 = count_1 + (i - prev - 1) / 2;` ` ` `prev = i;` ` ` `}` ` ` ` ` `// Check if prev and current are 1` ` ` `else` `if` `(S[prev] == ` `'1'` `&& S[i] == ` `'1'` `)` ` ` `{` ` ` ` ` `// All * will get converted to 1` ` ` `count_1 = count_1 + (i - prev - 1);` ` ` `prev = i;` ` ` `}` ` ` ` ` `// No * can be replaced` ` ` `// by either 0 or 1` ` ` `else` `if` `(S[prev] == ` `'0'` `&& S[i] == ` `'1'` `)` ` ` ` ` `// Prev becomes the ith character` ` ` `prev = i;` ` ` ` ` `// Check if prev and current are 0` ` ` `else` `if` `(S[prev] == ` `'0'` `&& S[i] == ` `'0'` `)` ` ` `{` ` ` ` ` `// All * will get converted to 0` ` ` `count_0 = count_0 + (i - prev - 1);` ` ` `prev = i;` ` ` `}` ` ` `}` ` ` ` ` `// If frequency of 0` ` ` `// is more` ` ` `if` `(count_0 > count_1)` ` ` `Console.Write(` `"0"` `);` ` ` ` ` `// If frequency of 1` ` ` `// is more` ` ` `else` `if` `(count_1 > count_0)` ` ` `Console.Write(` `"1"` `);` ` ` ` ` `else` ` ` `{` ` ` `Console.Write(` `"-1"` `);` ` ` `}` `}` ` ` `// Driver code` `public` `static` `void` `Main ()` `{` ` ` ` ` `// Given string` ` ` `string` `str = ` `"**0**1***0"` `;` ` ` ` ` `// Function call` ` ` `solve(str);` `}` `}` ` ` `// This code is contributed by code_hunt` |

**Output:**

0

**Time Complexity: **O(N) **Auxiliary Space: **O(1)

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