Given a binary array, we need to convert this array into an array that either contains all 1s or all 0s. We need to do it using the minimum number of group flips.

Examples :

Input: arr[] = {1, 1, 0, 0, 0, 1}Output: From 2 to 4Explanation: We have two choices, we make all 0s or do all 1s. We need to do two group flips to make all elements 0 and one group flip to make all elements 1. Since making all elements 1 takes least group flips, we do this.

Input: arr[] = {1, 0, 0, 0, 1, 0, 0, 1, 0, 1}Output:

From 1 to 3

From 5 to 6

From 8 to 8

Input: arr[] = {0, 0, 0}Output:Explanation: Output is empty, we need not to make any change

Input: arr[] = {1, 1, 1}Output:Explanation: Output is empty, we need not to make any change

Input: arr[] = {0, 1}Output:

From 0 to 0OR

From 1 to 1Explanation: Here number of flips are same either we make all elements as 1 or all elements as 0.

A **Naive Solution **is to traverse do two traversals of the array. We first traverse to find the number of groups of 0s and the number of groups of 1. We find the minimum of these two. Then we traverse the array and flip the 1s if groups of 1s are less. Otherwise we flip 0s.

**How to do it with one traversal of array?**

An **Efficient Solution **is based on the below facts :

- There are only two types of groups (groups of 0s and groups of 1s)
- Either the counts of both groups are same or the difference between counts is at most 1. For example, in {1, 1, 0, 1, 0, 0} there are two groups of 0s and two groups of 1s. In example, {1, 1, 0, 0, 0, 1, 0, 0, 1, 1}, count of groups of 1 is one more than the counts of 0s.

Based on the above facts, we can conclude that if we always flip the second group and other groups that of the same type as the second group, we always get the correct answer. In the first case, when group counts are the same, it does not matter which group type we flip as both will lead to the correct answer. In the second case, when there is one extra, by ignoring the first group and starting from the second group, we convert this case to first case (for subarray beginning from the second group) and get the correct answer.

## C++

`// C++ program to find the minimum ` `// group flips in a binary array ` `#include <iostream> ` `using` `namespace` `std; ` ` ` `void` `printGroups(` `bool` `arr[], ` `int` `n) { ` ` ` ` ` `// Traverse through all array elements ` ` ` `// starting from the second element ` ` ` `for` `(` `int` `i = 1; i < n; i++) { ` ` ` ` ` `// If current element is not same ` ` ` `// as previous ` ` ` `if` `(arr[i] != arr[i - 1]) { ` ` ` ` ` `// If it is same as first element ` ` ` `// then it is starting of the interval ` ` ` `// to be flipped. ` ` ` `if` `(arr[i] != arr[0]) ` ` ` `cout << ` `"From "` `<< i << ` `" to "` `; ` ` ` ` ` `// If it is not same as previous ` ` ` `// and same as first element, then ` ` ` `// previous element is end of interval ` ` ` `else` ` ` `cout << (i - 1) << endl; ` ` ` `} ` ` ` `} ` ` ` ` ` `// Explicitly handling the end of ` ` ` `// last interval ` ` ` `if` `(arr[n - 1] != arr[0]) ` ` ` `cout << (n - 1) << endl; ` `} ` ` ` `int` `main() { ` ` ` `bool` `arr[] = {0, 1, 1, 0, 0, 0, 1, 1}; ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` `printGroups(arr, n); ` ` ` `return` `0; ` `}` |

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## Java

`// Java program to find the minimum ` `// group flips in a binary array ` `import` `java.io.*; ` `import` `java.util.*; ` ` ` `class` `GFG { ` ` ` `static` `void` `printGroups(` `int` `arr[], ` `int` `n) ` `{ ` ` ` ` ` `// Traverse through all array elements ` ` ` `// starting from the second element ` ` ` `for` `(` `int` `i = ` `1` `; i < n; i++) ` ` ` `{ ` ` ` ` ` `// If current element is not same ` ` ` `// as previous ` ` ` `if` `(arr[i] != arr[i - ` `1` `]) ` ` ` `{ ` ` ` ` ` `// If it is same as first element ` ` ` `// then it is starting of the interval ` ` ` `// to be flipped. ` ` ` `if` `(arr[i] != arr[` `0` `]) ` ` ` `System.out.print(` `"From "` `+ i + ` `" to "` `); ` ` ` ` ` `// If it is not same as previous ` ` ` `// and same as first element, then ` ` ` `// previous element is end of interval ` ` ` `else` ` ` `System.out.println(i - ` `1` `); ` ` ` `} ` ` ` `} ` ` ` ` ` `// Explicitly handling the end of ` ` ` `// last interval ` ` ` `if` `(arr[n - ` `1` `] != arr[` `0` `]) ` ` ` `System.out.println(n - ` `1` `); ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` `int` `arr[] = {` `0` `, ` `1` `, ` `1` `, ` `0` `, ` `0` `, ` `0` `, ` `1` `, ` `1` `}; ` ` ` `int` `n = arr.length; ` ` ` ` ` `printGroups(arr, n); ` `} ` `} ` ` ` `// This code is contributed by coder001 ` |

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## Python3

`# Python3 program to find the minimum ` `# group flips in a binary array ` ` ` `def` `printGroups(arr, n): ` ` ` ` ` `# Traverse through all array elements ` ` ` `# starting from the second element ` ` ` `for` `i ` `in` `range` `(` `1` `, n): ` ` ` ` ` `# If current element is not same ` ` ` `# as previous ` ` ` `if` `(arr[i] !` `=` `arr[i ` `-` `1` `]): ` ` ` ` ` `# If it is same as first element ` ` ` `# then it is starting of the interval ` ` ` `# to be flipped. ` ` ` `if` `(arr[i] !` `=` `arr[` `0` `]): ` ` ` `print` `(` `"From"` `, i, ` `"to "` `, end ` `=` `"") ` ` ` ` ` `# If it is not same as previous ` ` ` `# and same as the first element, then ` ` ` `# previous element is end of interval ` ` ` `else` `: ` ` ` `print` `(i ` `-` `1` `) ` ` ` ` ` `# Explicitly handling the end of ` ` ` `# last interval ` ` ` `if` `(arr[n ` `-` `1` `] !` `=` `arr[` `0` `]): ` ` ` `print` `(n ` `-` `1` `) ` ` ` `# Driver Code ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` ` ` `arr ` `=` `[ ` `0` `, ` `1` `, ` `1` `, ` `0` `, ` `0` `, ` `0` `, ` `1` `, ` `1` `] ` ` ` `n ` `=` `len` `(arr) ` ` ` ` ` `printGroups(arr, n) ` ` ` `# This code is contributed by Bhupendra_Singh ` |

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## C#

`// C# program to find the minimum ` `// group flips in a binary array ` `using` `System; ` ` ` `class` `GFG{ ` ` ` `static` `void` `printGroups(` `int` `[]arr, ` `int` `n) ` `{ ` ` ` ` ` `// Traverse through all array elements ` ` ` `// starting from the second element ` ` ` `for` `(` `int` `i = 1; i < n; i++) ` ` ` `{ ` ` ` ` ` `// If current element is not same ` ` ` `// as previous ` ` ` `if` `(arr[i] != arr[i - 1]) ` ` ` `{ ` ` ` ` ` `// If it is same as first element ` ` ` `// then it is starting of the interval ` ` ` `// to be flipped. ` ` ` `if` `(arr[i] != arr[0]) ` ` ` `Console.Write(` `"From "` `+ i + ` `" to "` `); ` ` ` ` ` `// If it is not same as previous ` ` ` `// and same as first element, then ` ` ` `// previous element is end of interval ` ` ` `else` ` ` `Console.WriteLine(i - 1); ` ` ` `} ` ` ` `} ` ` ` ` ` `// Explicitly handling the end ` ` ` `// of last interval ` ` ` `if` `(arr[n - 1] != arr[0]) ` ` ` `Console.WriteLine(n - 1); ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` ` ` `int` `[]arr = { 0, 1, 1, 0, 0, 0, 1, 1 }; ` ` ` `int` `n = arr.Length; ` ` ` ` ` `printGroups(arr, n); ` `} ` `} ` ` ` `// This code is contributed by amal kumar choubey ` |

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**Output**

From 1 to 2 From 6 to 7

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

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