Minimum number of flips with rotation to make binary string alternating
Last Updated :
30 Nov, 2021
Given a binary string S of 0s and 1s. The task is to make the given string a sequence of alternate characters by using the below operations:
- Remove some prefix from the start and append it to the end.
- Flip some or every bit in the given string.
Print the minimum number of bits to be flipped to make the given string alternating.
Examples:
Input: S = "001"
Output: 0
Explanation:
No need to flip any element we can get alternating sequence by using left rotation: 010.
Input: S = "000001100"
Output: 3
Explanation:
Following steps to find minimum flips to get alternating string:
1. After rotating string 6 times towards left we will get: 100000001
2. Now we can apply flip operation as following: 101000001 -> 101010001 -> 101010101
Thus, minimum flips to make string alternating is 3.
Naive Approach: The naive approach is to take all N possible combinations and calculate the minimum number of bits To flip in each of those strings. Print the minimum count among all such combinations.
Time Complexity: O(N2), where N is the length of the string.
Auxiliary Space: O(N)
Efficient Approach: This can be solved by observing that the final string will be either of type "101010..." or "010101..." such that all 1s will either be at odd positions or at even positions. Follow the below steps to solve the problem:
- Create a prefix sum array where pref[i] means a number of changes required until index i.
- Create prefix arrays for both the above patterns.
- Check for every i, if substring[0, i] is appended at the end how many characters to be flipped required.
- Print the minimum number of flips among all the substrings in the above steps.
Why only odd length strings are considered for rotation and why rotation will have no effect on even length string?
I'll try to explain this with the below example,
Suppose you have the sequence 011000 which is of even length. Without rotation, you can change it to 010101 or 101010. Now suppose you choose to append 01 at the end. The sequence becomes 100001 which can be changed to 010101 or 101010. If you compare each character, you will see that this is the same case as that of without rotation. 1000 corresponds to either 0101 or 1010 in both cases and 01 to 01 or 10.
But now consider an odd length case, 01100. Without rotation, you can change it to 01010 or 10101. Now suppose you choose to append 01 at the end. The sequence becomes 10001 which can be changed to 01010 or 10101. Now if you compare each character, you will see that 100 corresponds to 010 or 101 in both cases but 01 corresponds to 01 when 100 is 010 in case of no rotation and 101 in case of rotation.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function that finds the minimum
// number of flips to make the
// binary string alternating if
// left circular rotation is allowed
int MinimumFlips(string s, int n)
{
int a[n];
for(int i = 0; i < n; i++)
{
a[i] = (s[i] == '1' ? 1 : 0);
}
// Initialize prefix arrays to store
// number of changes required to put
// 1s at either even or odd position
int oddone[n + 1];
int evenone[n + 1];
oddone[0] = 0;
evenone[0] = 0;
for(int i = 0; i < n; i++)
{
// If i is odd
if (i % 2 != 0)
{
// Update the oddone
// and evenone count
oddone[i + 1] = oddone[i] +
(a[i] == 1 ? 1 : 0);
evenone[i + 1] = evenone[i] +
(a[i] == 0 ? 1 : 0);
}
// Else i is even
else
{
// Update the oddone
// and evenone count
oddone[i + 1] = oddone[i] +
(a[i] == 0 ? 1 : 0);
evenone[i + 1] = evenone[i] +
(a[i] == 1 ? 1 : 0);
}
}
// Initialize minimum flips
int minimum = min(oddone[n], evenone[n]);
// Check if substring[0, i] is
// appended at end how many
// changes will be required
for(int i = 0; i < n; i++)
{
if (n % 2 != 0)
{
minimum = min(minimum,
oddone[n] -
oddone[i + 1] +
evenone[i + 1]);
minimum = min(minimum,
evenone[n] -
evenone[i + 1] +
oddone[i + 1]);
}
}
// Return minimum flips
return minimum;
}
// Driver Code
int main()
{
// Given String
string S = "000001100";
// Length of given string
int n = S.length();
// Function call
cout << (MinimumFlips(S, n));
}
// This code is contributed by chitranayal
// Java program for the above approach
import java.util.*;
class GFG {
// Function that finds the minimum
// number of flips to make the
// binary string alternating if
// left circular rotation is allowed
static int MinimumFlips(String s, int n)
{
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = (s.charAt(i) == '1' ? 1 : 0);
}
// Initialize prefix arrays to store
// number of changes required to put
// 1s at either even or odd position
int[] oddone = new int[n + 1];
int[] evenone = new int[n + 1];
oddone[0] = 0;
evenone[0] = 0;
for (int i = 0; i < n; i++) {
// If i is odd
if (i % 2 != 0) {
// Update the oddone
// and evenone count
oddone[i + 1]
= oddone[i]
+ (a[i] == 1 ? 1 : 0);
evenone[i + 1]
= evenone[i]
+ (a[i] == 0 ? 1 : 0);
}
// Else i is even
else {
// Update the oddone
// and evenone count
oddone[i + 1]
= oddone[i]
+ (a[i] == 0 ? 1 : 0);
evenone[i + 1]
= evenone[i]
+ (a[i] == 1 ? 1 : 0);
}
}
// Initialize minimum flips
int minimum = Math.min(oddone[n],
evenone[n]);
// Check if substring[0, i] is
// appended at end how many
// changes will be required
for (int i = 0; i < n; i++) {
if (n % 2 != 0) {
minimum = Math.min(minimum,
oddone[n]
- oddone[i + 1]
+ evenone[i + 1]);
minimum = Math.min(minimum,
evenone[n]
- evenone[i + 1]
+ oddone[i + 1]);
}
}
// Return minimum flips
return minimum;
}
// Driver Code
public static void main(String[] args)
{
// Given String
String S = "000001100";
// Length of given string
int n = S.length();
// Function call
System.out.print(MinimumFlips(S, n));
}
}
# Python3 program for the above approach
# Function that finds the minimum
# number of flips to make the
# binary string alternating if
# left circular rotation is allowed
def MinimumFlips(s, n):
a = [0] * n
for i in range(n):
a[i] = 1 if s[i] == '1' else 0
# Initialize prefix arrays to store
# number of changes required to put
# 1s at either even or odd position
oddone = [0] * (n + 1)
evenone = [0] * (n + 1)
for i in range(n):
# If i is odd
if(i % 2 != 0):
# Update the oddone
# and evenone count
if(a[i] == 1):
oddone[i + 1] = oddone[i] + 1
else:
oddone[i + 1] = oddone[i] + 0
if(a[i] == 0):
evenone[i + 1] = evenone[i] + 1
else:
evenone[i + 1] = evenone[i] + 0
# Else i is even
else:
# Update the oddone
# and evenone count
if (a[i] == 0):
oddone[i + 1] = oddone[i] + 1
else:
oddone[i + 1] = oddone[i] + 0
if (a[i] == 1):
evenone[i + 1] = evenone[i] + 1
else:
evenone[i + 1] = evenone[i] + 0
# Initialize minimum flips
minimum = min(oddone[n], evenone[n])
# Check if substring[0, i] is
# appended at end how many
# changes will be required
for i in range(n):
if(n % 2 != 0):
minimum = min(minimum,
oddone[n] -
oddone[i + 1] +
evenone[i + 1])
minimum = min(minimum,
evenone[n] -
evenone[i + 1] +
oddone[i + 1])
# Return minimum flips
return minimum
# Driver Code
# Given String
S = "000001100"
# Length of given string
n = len(S)
# Function call
print(MinimumFlips(S, n))
# This code is contributed by Shivam Singh
// C# program for the above approach
using System;
class GFG{
// Function that finds the minimum
// number of flips to make the
// binary string alternating if
// left circular rotation is allowed
static int MinimumFlips(String s, int n)
{
int[] a = new int[n];
for (int i = 0; i < n; i++)
{
a[i] = (s[i] == '1' ? 1 : 0);
}
// Initialize prefix arrays to store
// number of changes required to put
// 1s at either even or odd position
int[] oddone = new int[n + 1];
int[] evenone = new int[n + 1];
oddone[0] = 0;
evenone[0] = 0;
for (int i = 0; i < n; i++)
{
// If i is odd
if (i % 2 != 0)
{
// Update the oddone
// and evenone count
oddone[i + 1] = oddone[i] +
(a[i] == 1 ? 1 : 0);
evenone[i + 1] = evenone[i] +
(a[i] == 0 ? 1 : 0);
}
// Else i is even
else
{
// Update the oddone
// and evenone count
oddone[i + 1] = oddone[i] +
(a[i] == 0 ? 1 : 0);
evenone[i + 1] = evenone[i] +
(a[i] == 1 ? 1 : 0);
}
}
// Initialize minimum flips
int minimum = Math.Min(oddone[n],
evenone[n]);
// Check if substring[0, i] is
// appended at end how many
// changes will be required
for (int i = 0; i < n; i++)
{
if (n % 2 != 0)
{
minimum = Math.Min(minimum, oddone[n] -
oddone[i + 1] +
evenone[i + 1]);
minimum = Math.Min(minimum, evenone[n] -
evenone[i + 1] +
oddone[i + 1]);
}
}
// Return minimum flips
return minimum;
}
// Driver Code
public static void Main(String[] args)
{
// Given String
String S = "000001100";
// Length of given string
int n = S.Length;
// Function call
Console.Write(MinimumFlips(S, n));
}
}
// This code is contributed by Rajput-Ji
<script>
// JavaScript program for the
// above approach
// Function that finds the minimum
// number of flips to make the
// binary string alternating if
// left circular rotation is allowed
function MinimumFlips(s, n)
{
let a = Array.from({length: n+1}, (_, i) => 0);
for (let i = 0; i < n; i++) {
a[i] = (s[i] == '1' ? 1 : 0);
}
// Initialize prefix arrays to store
// number of changes required to put
// 1s at either even or odd position
let oddone = Array.from({length: n+1}, (_, i) => 0);
let evenone = Array.from({length: n+1}, (_, i) => 0);
oddone[0] = 0;
evenone[0] = 0;
for (let i = 0; i < n; i++) {
// If i is odd
if (i % 2 != 0) {
// Update the oddone
// and evenone count
oddone[i + 1]
= oddone[i]
+ (a[i] == 1 ? 1 : 0);
evenone[i + 1]
= evenone[i]
+ (a[i] == 0 ? 1 : 0);
}
// Else i is even
else {
// Update the oddone
// and evenone count
oddone[i + 1]
= oddone[i]
+ (a[i] == 0 ? 1 : 0);
evenone[i + 1]
= evenone[i]
+ (a[i] == 1 ? 1 : 0);
}
}
// Initialize minimum flips
let minimum = Math.min(oddone[n],
evenone[n]);
// Check if substring[0, i] is
// appended at end how many
// changes will be required
for (let i = 0; i < n; i++) {
if (n % 2 != 0) {
minimum = Math.min(minimum,
oddone[n]
- oddone[i + 1]
+ evenone[i + 1]);
minimum = Math.min(minimum,
evenone[n]
- evenone[i + 1]
+ oddone[i + 1]);
}
}
// Return minimum flips
return minimum;
}
// Driver Code
// Given String
let S = "000001100";
// Length of given string
let n = S.length;
// Function call
document.write(MinimumFlips(S, n));
</script>
Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(N)
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