Minimum number of flips or swaps of adjacent characters required to make two strings equal
Last Updated :
23 Jul, 2025
Given two binary strings A and B of length N, the task is to count the minimum number of operations required to make the two given strings equal by either swapping adjacent characters or flipping any character of the string A.
Examples:
Input: A = "100", B = "001"
Output: 2
Explanation: Flipping characters A[0](= '1') and A[2](= '0') modifies the string A to "001", which is equal to the string B.
Input: A = "0101", B = "0011"
Output: 1
Explanation: Swapping the characters A[2](= '0') and A[3](= '1') modifies the string A to "0011", which is equal to string B.
Approach: The problem can be solved using Dynamic Programming as it has Overlapping Subproblems and Optimal Substructure.
Follow the steps below to solve the problem:
- Initialize an array, say dp[] of size N+1 as {0}, where dp[i] stores the minimum number of operations required up to index i, to make the prefix of Ai equal to the prefix Bi.
- Iterate over the range [1, N] using a variable, say i, and performing the following operations:
- If A[i - 1] equals B[i - 1], then update dp[i] to dp[i - 1].
- Otherwise, update dp[i] to dp[i - 1] + 1.
- If swapping is possible, i.e. i > 1 and A[i - 2] is equal to B[i - 1] and A[i - 1] is equal to B[i - 2], then update dp[i] to min(dp[i], dp[i - 2] + 1).
- After completing the above steps, print the minimum number of operations obtained, i.e. the value dp[N].
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count the minimum
// number of operations required
// to make strings A and B equal
int countMinSteps(string A, string B, int N)
{
// Stores all dp-states
vector<int> dp(N + 1, 0);
// Iterate over the range [1, N]
for (int i = 1; i <= N; i++) {
// If A[i - 1] equals to B[i - 1]
if (A[i - 1] == B[i - 1]) {
// Assign Dp[i - 1] to Dp[i]
dp[i] = dp[i - 1];
}
// Otherwise
else {
// Update dp[i]
dp[i] = dp[i - 1] + 1;
}
// If swapping is possible
if (i >= 2 && A[i - 2] == B[i - 1]
&& A[i - 1] == B[i - 2]) {
// Update dp[i]
dp[i] = min(dp[i], dp[i - 2] + 1);
}
}
// Return the minimum
// number of steps required
return dp[N];
}
// Driver Code
int main()
{
// Given Input
string A = "0101";
string B = "0011";
int N = A.length();
// Function Call
cout << countMinSteps(A, B, N);
return 0;
}
// Java program for the above approach
import java.io.*;
class GFG{
// Function to count the minimum
// number of operations required
// to make strings A and B equal
static int countMinSteps(String A, String B, int N)
{
// Stores all dp-states
int[] dp = new int[N + 1];
for(int i = 1; i <= N; i++)
{
// Update the value of A[i]
dp[i] = 0;
}
// Iterate over the range [1, N]
for(int i = 1; i <= N; i++)
{
// If A[i - 1] equals to B[i - 1]
if (A.charAt(i - 1) == B.charAt(i - 1))
{
// Assign Dp[i - 1] to Dp[i]
dp[i] = dp[i - 1];
}
// Otherwise
else
{
// Update dp[i]
dp[i] = dp[i - 1] + 1;
}
// If swapping is possible
if (i >= 2 && A.charAt(i - 2) == B.charAt(i - 1) &&
A.charAt(i - 1) == B.charAt(i - 2))
{
// Update dp[i]
dp[i] = Math.min(dp[i], dp[i - 2] + 1);
}
}
// Return the minimum
// number of steps required
return dp[N];
}
// Driver code
public static void main(String[] args)
{
// Given Input
String A = "0101";
String B = "0011";
int N = A.length();
// Function Call
System.out.println(countMinSteps(A, B, N));
}
}
// This code is contributed by sanjoy_62
# Python3 program for the above approach
# Function to count the minimum
# number of operations required
# to make strings A and B equal
def countMinSteps(A, B, N) :
# Stores all dp-states
dp = [0] * (N + 1)
# Iterate rate over the range [1, N]
for i in range(1, N+1) :
# If A[i - 1] equals to B[i - 1]
if (A[i - 1] == B[i - 1]) :
# Assign Dp[i - 1] to Dp[i]
dp[i] = dp[i - 1]
# Otherwise
else :
# Update dp[i]
dp[i] = dp[i - 1] + 1
# If swapping is possible
if (i >= 2 and A[i - 2] == B[i - 1]
and A[i - 1] == B[i - 2]) :
# Update dp[i]
dp[i] = min(dp[i], dp[i - 2] + 1)
# Return the minimum
# number of steps required
return dp[N]
# Driver Code
# Given Input
A = "0101"
B = "0011"
N = len(A)
# Function Call
print(countMinSteps(A, B, N))
# This code is contributed by splevel62.
// C# program for the above approach
using System;
class GFG{
// Function to count the minimum
// number of operations required
// to make strings A and B equal
static int countMinSteps(string A, string B, int N)
{
// Stores all dp-states
int[] dp = new int[N + 1];
for(int i = 1; i <= N; i++)
{
// Update the value of A[i]
dp[i] = 0;
}
// Iterate over the range [1, N]
for(int i = 1; i <= N; i++)
{
// If A[i - 1] equals to B[i - 1]
if (A[i - 1] == B[i - 1])
{
// Assign Dp[i - 1] to Dp[i]
dp[i] = dp[i - 1];
}
// Otherwise
else
{
// Update dp[i]
dp[i] = dp[i - 1] + 1;
}
// If swapping is possible
if (i >= 2 && A[i - 2] == B[i - 1] &&
A[i - 1] == B[i - 2])
{
// Update dp[i]
dp[i] = Math.Min(dp[i], dp[i - 2] + 1);
}
}
// Return the minimum
// number of steps required
return dp[N];
}
// Driver code
public static void Main(String []args)
{
// Given Input
string A = "0101";
string B = "0011";
int N = A.Length;
// Function Call
Console.Write(countMinSteps(A, B, N));
}
}
// This code is contributed by code_hunt
<script>
// JavaScript Program for the above approach
// Function to count the minimum
// number of operations required
// to make strings A and B equal
function countMinSteps(A, B, N) {
// Stores all dp-states
let dp = new Array(N + 1).fill(0);
// Iterate over the range [1, N]
for (let i = 1; i <= N; i++) {
// If A[i - 1] equals to B[i - 1]
if (A[i - 1] == B[i - 1]) {
// Assign Dp[i - 1] to Dp[i]
dp[i] = dp[i - 1];
}
// Otherwise
else {
// Update dp[i]
dp[i] = dp[i - 1] + 1;
}
// If swapping is possible
if (i >= 2 && A[i - 2] == B[i - 1]
&& A[i - 1] == B[i - 2]) {
// Update dp[i]
dp[i] = Math.min(dp[i], dp[i - 2] + 1);
}
}
// Return the minimum
// number of steps required
return dp[N];
}
// Driver Code
// Given Input
let A = "0101";
let B = "0011";
let N = A.length;
// Function Call
document.write(countMinSteps(A, B, N));
// This code is contributed by Potta Lokesh
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)
Efficient approach : Space optimization O(1)
In previous approach we the current value dp[i] is only depend upon the previous 2 values i.e. dp[i-1] and dp[i-2]. So to optimize the space we can keep track of previous and current values by the help of three variables prev1, prev2 and curr which will reduce the space complexity from O(N) to O(1).
Implementation Steps:
- Create 2 variables prev1 and prev2 to keep track o previous values of DP.
- Initialize base case prev1 = prev2 = 0.
- Create a variable curr to store current value.
- Iterate over subproblem using loop and update curr.
- After every iteration update prev1 and prev2 for further iterations.
- At last return curr.
Implementation:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count the minimum
// number of operations required
// to make strings A and B equal
int countMinSteps(string A, string B, int N)
{
// variables to Stores all dp-states
int prev1 = 0, prev2 = 0, curr = 0;
// Iterate over the range [1, N]
for (int i = 1; i <= N; i++) {
// If A[i - 1] equals to B[i - 1]
if (A[i - 1] == B[i - 1]){
// Assign prev1 to curr
curr = prev1;
}
else{
// Update dp[i]
curr = prev1 + 1;
}
// If swapping is possible
if (i >= 2 && A[i - 2] == B[i - 1] && A[i - 1] == B[i - 2]){
// Update curr
curr = min(curr, prev2 + 1);
}
// assigning values for further iteration
prev2 = prev1;
prev1 = curr;
}
// return answer
return curr;
}
// Driver Code
int main()
{
string A = "0101";
string B = "0011";
int N = A.length();
// function call
cout << countMinSteps(A, B, N);
return 0;
}
// --- by bhardwajji
import java.util.*;
public class Main
{
// Function to count the minimum
// number of operations required
// to make strings A and B equal
static int countMinSteps(String A, String B, int N)
{
// variables to Stores all dp-states
int prev1 = 0, prev2 = 0, curr = 0;
// Iterate over the range [1, N]
for (int i = 1; i <= N; i++) {
// If A[i - 1] equals to B[i - 1]
if (A.charAt(i - 1) == B.charAt(i - 1)) {
// Assign prev1 to curr
curr = prev1;
} else {
// Update dp[i]
curr = prev1 + 1;
}
// If swapping is possible
if (i >= 2 && A.charAt(i - 2) == B.charAt(i - 1) && A.charAt(i - 1) == B.charAt(i - 2)) {
// Update curr
curr = Math.min(curr, prev2 + 1);
}
// assigning values for further iteration
prev2 = prev1;
prev1 = curr;
}
// return answer
return curr;
}
// Driver Code
public static void main(String[] args) {
String A = "0101";
String B = "0011";
int N = A.length();
// function call
System.out.println(countMinSteps(A, B, N));
}
}
def countMinSteps(A, B, N):
# variables to Stores all dp-states
prev1, prev2, curr = 0, 0, 0
# Iterate over the range [1, N]
for i in range(1, N+1):
# If A[i - 1] equals to B[i - 1]
if A[i - 1] == B[i - 1]:
# Assign prev1 to curr
curr = prev1
else:
# Update dp[i]
curr = prev1 + 1
# If swapping is possible
if i >= 2 and A[i - 2] == B[i - 1] and A[i - 1] == B[i - 2]:
# Update curr
curr = min(curr, prev2 + 1)
# assigning values for further iteration
prev2, prev1 = prev1, curr
# return answer
return curr
# Driver Code
A = "0101"
B = "0011"
N = len(A)
# function call
print(countMinSteps(A, B, N))
// C# program for the above approach
using System;
public class GFG {
// Function to count the minimum
// number of operations required
// to make strings A and B equal
public static int CountMinSteps(string A, string B,
int N)
{
// Variables to store all dp-states
int prev1 = 0, prev2 = 0, curr = 0;
// Iterate over the range [1, N]
for (int i = 1; i <= N; i++) {
// If A[i - 1] equals to B[i - 1]
if (A[i - 1] == B[i - 1]) {
// Assign prev1 to curr
curr = prev1;
}
else {
// Update dp[i]
curr = prev1 + 1;
}
// If swapping is possible
if (i >= 2 && A[i - 2] == B[i - 1]
&& A[i - 1] == B[i - 2]) {
// Update curr
curr = Math.Min(curr, prev2 + 1);
}
// Assigning values for further iteration
prev2 = prev1;
prev1 = curr;
}
// Return the answer
return curr;
}
// Driver Code
public static void Main(string[] args)
{
string A = "0101";
string B = "0011";
int N = A.Length;
// Function call
Console.WriteLine(CountMinSteps(A, B, N));
}
}
// This code is contributed by Susobhan Akhuli
// JavaScript program for the above approach
// Function to count the minimum number of operations required
// to make strings A and B equal
function countMinSteps(A, B, N) {
// Variables to store all dp states
let prev1 = 0;
let prev2 = 0;
let curr = 0;
// Iterate over the range [1, N]
for (let i = 1; i <= N; i++) {
// If A[i - 1] equals B[i - 1]
if (A[i - 1] === B[i - 1]) {
// Assign prev1 to curr
curr = prev1;
} else {
// Update dp[i]
curr = prev1 + 1;
}
// If swapping is possible
if (i >= 2 && A[i - 2] === B[i - 1] && A[i - 1] === B[i - 2]) {
// Update curr
curr = Math.min(curr, prev2 + 1);
}
// Assigning values for further iteration
prev2 = prev1;
prev1 = curr;
}
// Return the answer
return curr;
}
// Driver Code
const A = "0101";
const B = "0011";
const N = A.length;
// Function call
document.write(countMinSteps(A, B, N));
// This code is contributed by Susobhan Akhuli
Output:
1
Time Complexity: O(N)
Auxiliary Space: O(1)
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