Minimize cost of swapping set bits with unset bits in a given Binary string
Last Updated :
26 Jul, 2025
Given a binary string S of size N, the task is to find the minimum cost by swapping every set bit with an unset bit such that the cost of swapping pairs of bits at indices i and j is abs(j - i).
Note: A swapped bit can't be swapped twice and the count of set bit in the given binary string is at most N/2.
Examples:
Input: S = "1010001"
Output: 3
Explanation:
Following the swapping of characters required:
- Swap characters at indices (0, 1) modifies the string to "0110001" and the cost of this operation is |1 - 0| = 1.
- Swap characters at indices (2, 3) modifies the string to "0101001" and the cost of this operation is |2 - 1| = 1.
- Swap characters at indices (6, 7) modifies the string to "0101010" and the cost of this operation is |7 - 6| = 1.
After the above operations, all the set bits is replaced with unset bits and the total cost of operations is 1 + 1 + 1 = 3.
Input: S = "1100"
Output: 4
Approach: The given problem can be solved using Dynamic Programming by storing the indices of set and unset bits in two auxiliary arrays, say A[] and B[], and then find the sum of the difference between array elements A[] with any element of the array B[]. Follow the steps below to solve the given problem:
- Initialize two arrays, say A[] and B[], and store the indices of set and unset bits in it.
- Initialize a 2D array, dp[][] of dimensions K*(N - K) where K is the count of set bit in S such thatdp[i][j] stores the minimum cost of swapping the ith array element A[] with the jth array element B[].
- Now, for each state there are two choices:
- Swap the ith array element A[] till the (j - 1)th array element B[] as dp[i][j] = dp[i][j - 1].
- Swap the (i - 1)th array element A[] till the (j - 1)th array element B[] and the ith array element A[] with jth array element B[] and this state can be calculated as dp[i][j] = dp[i - 1][j - 1] + abs(A[i] - B[i]).
- Now, choose the minimum of the above two choices to find the current state as:
dp[i][j] = min(dp[i][j-1], dp[i-1][j-1] + abs(A[i] - B[j]))
- After completing the above steps, print the value of dp[K][N - K] as the resultant minimum number of operations.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
#define INF 1000000000
// Function to find the minimum cost
// required to swap every set bit with
// an unset bit
int minimumCost(string s)
{
int N = s.length();
// Stores the indices of set and
// unset bits of the string S
vector<int> A, B;
// Traverse the string S
for (int i = 0; i < N; i++) {
// Store the indices
if (s[i] == '1') {
A.push_back(i);
}
else {
B.push_back(i);
}
}
int n1 = A.size();
int n2 = B.size();
// Initialize a dp table of size
// n1*n2
int dp[n1 + 1][n2 + 1];
// Initialize all states to 0
memset(dp, 0, sizeof(dp));
// Set unreachable states to INF
for (int i = 1; i <= n1; i++) {
dp[i][0] = INF;
}
// Fill the dp Table according to
// the given recurrence relation
for (int i = 1; i <= n1; i++) {
for (int j = 1; j <= n2; j++) {
// Update the value of
// dp[i][j]
dp[i][j] = min(
dp[i][j - 1],
dp[i - 1][j - 1]
+ abs(A[i - 1] - B[j - 1]));
}
}
// Return the minimum cost
return dp[n1][n2];
}
// Driver Code
int main()
{
string S = "1010001";
cout << minimumCost(S);
return 0;
}
// Java program for the above approach
import java.util.*;
class GFG{
static final int INF = 1000000000;
// Function to find the minimum cost
// required to swap every set bit with
// an unset bit
static int minimumCost(String s)
{
int N = s.length();
// Stores the indices of set and
// unset bits of the String S
Vector<Integer> A = new Vector<Integer>();
Vector<Integer> B = new Vector<Integer>();
// Traverse the String S
for (int i = 0; i < N; i++) {
// Store the indices
if (s.charAt(i) == '1') {
A.add(i);
}
else {
B.add(i);
}
}
int n1 = A.size();
int n2 = B.size();
// Initialize a dp table of size
// n1*n2
int [][]dp = new int[n1 + 1][n2 + 1];
// Set unreachable states to INF
for (int i = 1; i <= n1; i++) {
dp[i][0] = INF;
}
// Fill the dp Table according to
// the given recurrence relation
for (int i = 1; i <= n1; i++) {
for (int j = 1; j <= n2; j++) {
// Update the value of
// dp[i][j]
dp[i][j] = Math.min(
dp[i][j - 1],
dp[i - 1][j - 1]
+ Math.abs(A.get(i - 1) - B.get(j - 1)));
}
}
// Return the minimum cost
return dp[n1][n2];
}
// Driver Code
public static void main(String[] args)
{
String S = "1010001";
System.out.print(minimumCost(S));
}
}
// This code is contributed by shikhasingrajput
# Python program for the above approach
INF = 1000000000
# Function to find the minimum cost
# required to swap every set bit with
# an unset bit
def minimumCost(s):
N = len(s)
# Stores the indices of set and
# unset bits of the string S
A = []
B = []
# Traverse the string S
for i in range(0, N):
# Store the indices
if (s[i] == "1"):
A.append(i)
else:
B.append(i)
n1 = len(A)
n2 = len(B)
# Initialize a dp table of size
# n1*n2
dp = [[0 for i in range(n2 + 1)] for j in range(n1 + 1)]
# Set unreachable states to INF
for i in range(1, n1 + 1):
dp[i][0] = INF
# Fill the dp Table according to
# the given recurrence relation
for i in range(1, n1 + 1):
for j in range(1, n2 + 1):
# Update the value of
# dp[i][j]
dp[i][j] = min(
dp[i][j - 1],
dp[i - 1][j - 1] + abs(A[i - 1] - B[j - 1])
)
# Return the minimum cost
return dp[n1][n2]
# Driver Code
S = "1010001"
print(minimumCost(S))
# This code is contributed by _saurabh_jaiswal.
// C# program for the above approach
using System;
using System.Collections;
using System.Collections.Generic;
public class Program
{
// Function to find the minimum cost
// required to swap every set bit with
// an unset bit
static int minimumCost(string s)
{
int INF = 1000000000;
int N = s.Length;
// Stores the indices of set and
// unset bits of the string S
List<int> A = new List<int>();
List<int> B = new List<int>();
// Traverse the string S
for (int i = 0; i < N; i++) {
// Store the indices
if (s[i] == '1') {
A.Add(i);
}
else {
B.Add(i);
}
}
int n1 = A.Count;
int n2 = B.Count;
// Initialize a dp table of size
// n1*n2
int [,]dp = new int[n1 + 1,n2 + 1];
// Set unreachable states to INF
for (int i = 1; i <= n1; i++) {
dp[i,0] = INF;
}
// Fill the dp Table according to
// the given recurrence relation
for (int i = 1; i <= n1; i++) {
for (int j = 1; j <= n2; j++) {
// Update the value of
// dp[i][j]
dp[i,j] = Math.Min(
dp[i,j - 1],
dp[i - 1,j - 1]
+ Math.Abs(A[i - 1] - B[j - 1]));
}
}
// Return the minimum cost
return dp[n1,n2];
}
public static void Main()
{
string S = "1010001";
Console.Write(minimumCost(S));
}
}
// This code is contributed by rutvik_56.
<script>
// Javascript program for the above approach
let INF = 1000000000;
// Function to find the minimum cost
// required to swap every set bit with
// an unset bit
function minimumCost(s) {
let N = s.length;
// Stores the indices of set and
// unset bits of the string S
let A = [],
B = [];
// Traverse the string S
for (let i = 0; i < N; i++) {
// Store the indices
if (s[i] == "1") {
A.push(i);
} else {
B.push(i);
}
}
let n1 = A.length;
let n2 = B.length;
// Initialize a dp table of size
// n1*n2
let dp = new Array(n1 + 1).fill(0).map(() => new Array(n2 + 1).fill(0));
// Set unreachable states to INF
for (let i = 1; i <= n1; i++) {
dp[i][0] = INF;
}
// Fill the dp Table according to
// the given recurrence relation
for (let i = 1; i <= n1; i++) {
for (let j = 1; j <= n2; j++) {
// Update the value of
// dp[i][j]
dp[i][j] = Math.min(
dp[i][j - 1],
dp[i - 1][j - 1] + Math.abs(A[i - 1] - B[j - 1])
);
}
}
// Return the minimum cost
return dp[n1][n2];
}
// Driver Code
let S = "1010001";
document.write(minimumCost(S));
// This code is contributed by gfgking.
</script>
Time Complexity: O(K*(N - K)) where K is the count of set bit in S.
Auxiliary Space: O(K*(N - K))
Efficient approach : Space optimization
In previous approach the current value dp[i][j] is only depend upon the current and previous row values of DP. So to optimize the space complexity we use a single 1D array to store the computations.
Implementation steps:
- Create a 1D vector dp of size n+1 and initialize it with 0.
- Set a base case by initializing the values of DP .
- Now iterate over subproblems by the help of nested loop and get the current value from previous computations.
- At last return and print the final answer stored in dp[n].
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
#define INF 1000000000
// Function to find the minimum cost
// required to swap every set bit with
// an unset bit
int minimumCost(string s)
{
int n1 = 0, n2 = 0;
for (char c : s) {
if (c == '1') n1++;
else n2++;
}
// Initialize a dp table of size
// n1*n2
int dp[n1 + 1];
// Initialize all states to 0
memset(dp, 0, sizeof(dp));
// Set unreachable states to INF
for (int i = 1; i <= n1; i++) {
dp[i] = INF;
}
// Fill the dp Table according to
// the given recurrence relation
for (char c : s) {
for (int i = n1; i >= 1; i--) {
if (c == '0') dp[i] = min(dp[i], dp[i - 1] + n2 - i + 1);
else dp[i] = min(dp[i], dp[i - 1] + i - 1);
}
}
// Return the minimum cost
return dp[n1];
}
// Driver Code
int main()
{
string S = "1010001";
cout << minimumCost(S);
return 0;
}
import java.util.Arrays;
public class MinimumCost {
static int INF = 1000000000;
// Function to find the minimum cost
// required to swap every set bit with
// an unset bit
static int minimumCost(String s) {
int n1 = 0, n2 = 0;
for (char c : s.toCharArray()) {
if (c == '1') {
n1++;
} else {
n2++;
}
}
// Initialize a dp table of size
// n1*n2
int[] dp = new int[n1 + 1];
// Initialize all states to 0
Arrays.fill(dp, 0);
// Set unreachable states to INF
for (int i = 1; i <= n1; i++) {
dp[i] = INF;
}
// Fill the dp Table according to
// the given recurrence relation
for (char c : s.toCharArray()) {
for (int i = n1; i >= 1; i--) {
if (c == '0') {
dp[i] = Math.min(dp[i], dp[i - 1] + n2 - i + 1);
} else {
dp[i] = Math.min(dp[i], dp[i - 1] + i - 1);
}
}
}
// Return the minimum cost
return dp[n1];
}
// Driver Code
public static void main(String[] args) {
String S = "1010001";
System.out.println(minimumCost(S));
}
}
INF = 1000000000
# Function to find the minimum cost
# required to swap every set bit with an unset bit
def minimumCost(s):
n1 = s.count('1')
n2 = len(s) - n1
# Initialize a dp table of size n1*n2
dp = [0]*(n1 + 1)
# Set unreachable states to INF
for i in range(1, n1 + 1):
dp[i] = INF
# Fill the dp Table according
# to the given recurrence relation
for c in s:
for i in range(n1, 0, -1):
if c == '0':
dp[i] = min(dp[i], dp[i - 1] + n2 - i + 1)
else:
dp[i] = min(dp[i], dp[i - 1] + i - 1)
# Return the minimum cost
return dp[n1]
# Driver Code
if __name__ == "__main__":
S = "1010001"
print(minimumCost(S))
using System;
namespace MinimumCostToSwapBits
{
class Program
{
static int INF = 1000000000;
static int MinimumCost(string s)
{
int n1 = 0, n2 = 0;
foreach (char c in s)
{
if (c == '1') n1++;
else n2++;
}
int[] dp = new int[n1 + 1];
Array.Fill(dp, 0);
for (int i = 1; i <= n1; i++)
{
dp[i] = INF;
}
foreach (char c in s)
{
for (int i = n1; i >= 1; i--)
{
if (c == '0') dp[i] = Math.Min(dp[i], dp[i - 1] + n2 - i + 1);
else dp[i] = Math.Min(dp[i], dp[i - 1] + i - 1);
}
}
return dp[n1];
}
static void Main(string[] args)
{
string S = "1010001";
Console.WriteLine(MinimumCost(S));
}
}
}
function minimumCost(s) {
const INF = 1000000000;
let n1 = 0, n2 = 0;
for (let i = 0; i < s.length; i++) {
if (s.charAt(i) == '1') n1++;
else n2++;
}
const dp = new Array(n1 + 1).fill(0);
for (let i = 1; i <= n1; i++) {
dp[i] = INF;
}
for (let i = 0; i < s.length; i++) {
const c = s.charAt(i);
for (let j = n1; j >= 1; j--) {
if (c == '0') dp[j] = Math.min(dp[j], dp[j - 1] + n2 - j + 1);
else dp[j] = Math.min(dp[j], dp[j - 1] + j - 1);
}
}
return dp[n1];
}
const S = "1010001";
console.log(minimumCost(S)); // Output: 3
Time Complexity: O(N^2)
Auxiliary Space: O(N)
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