Minimum number of swaps required such that a given substring consists of exactly K 1s
Last Updated :
23 Jul, 2025
Given a binary string S of size N and three positive integers L, R, and K, the task is to find the minimum number of swaps required to such that the substring {S[L], .. S[R]} consists of exactly K 1s. If it is not possible to do so, then print "-1".
Examples:
Input: S = "110011111000101", L = 5, R = 8, K = 2
Output: 2
Explanation:
Initially, the substring {S[5], .. S[8]} = "1111" consists of 4 1s.
Swap (S[5], S[3]) and (S[6], S[4]).
Modified string S = "111100111000101" and {S[5], .. S[8]} = "0011".
Therefore, 2 swaps are required.
Input: S = "01010101010101", L = 3, R = 7, K = 8
Output: -1
Approach: The idea is to count the number of 1s and 0s present in the substring and outside the substring, {S[L], ..., S[R]}. Then, check if enough 1s or 0s are present outside that range which can be swapped such that the substring contains exactly K 1s.
Follow the steps below to solve the problem:
- Store the frequency of 1s and 0s in the string S in cntOnes and cntZeros respectively.
- Also, store the frequency of 1s and 0s in the substring S[L, R] in ones and zeros respectively.
- Find the frequency of 1s and 0s outside the substring S[L, R] using the formula: (rem_ones = cntOnes - ones) and rem_zero = (cntZeros - zeros).
- If k ? ones, then do the following:
- Initialize rem = (K - ones), which denotes the number of 1s required to make the sum equal to K.
- If zeros ? rem and rem_ones ? rem, print the value of rem as the result.
- Otherwise, if K < ones, then do the following:
- Initialize rem = (ones - K), which denotes the number of 0s required to make the sum equal to K.
- If ones ? rem and rem_zeros ? rem, print the value of rem as the result.
- In all other cases, print -1.
Below is the implementation of the above approach:
C++
// CPP program for the above approach
#include<bits/stdc++.h>
using namespace std;
// Function to find the minimum number
// of swaps required such that the
// substring {s[l], .., s[r]} consists
// of exactly k 1s
int minimumSwaps(string s, int l, int r, int k)
{
// Store the size of the string
int n = s.length();
// Store the total number of 1s
// and 0s in the entire string
int tot_ones = 0, tot_zeros = 0;
// Traverse the string S to find
// the frequency of 1 and 0
for (int i = 0; i < s.length(); i++)
{
if (s[i] == '1')
tot_ones++;
else
tot_zeros++;
}
// Store the number of 1s and
// 0s in the substring s[l, r]
int ones = 0, zeros = 0, sum = 0;
// Traverse the substring S[l, r]
// to find the frequency of 1s
// and 0s in it
for (int i = l - 1; i < r; i++)
{
if (s[i] == '1')
{
ones++;
sum++;
}
else
zeros++;
}
// Store the count of 1s and
// 0s outside substring s[l, r]
int rem_ones = tot_ones - ones;
int rem_zeros = tot_zeros - zeros;
// Check if the sum of the
// substring is at most K
if (k >= sum)
{
// Store number of 1s required
int rem = k - sum;
// Check if there are enough 1s
// remaining to be swapped
if (zeros >= rem && rem_ones >= rem)
return rem;
}
// If the count of 1s in the substring exceeds k
else if (k < sum)
{
// Store the number of 0s required
int rem = sum - k;
// Check if there are enough 0s
// remaining to be swapped
if (ones >= rem && rem_zeros >= rem)
return rem;
}
// In all other cases, print -1
return -1;
}
// Driver Code
int main()
{
string S = "110011111000101";
int L = 5, R = 8, K = 2;
cout<<minimumSwaps(S, L, R, K);
}
// This code is contributed bgangwar59.
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to find the minimum number
// of swaps required such that the
// substring {s[l], .., s[r]} consists
// of exactly k 1s
static int minimumSwaps(
String s, int l, int r, int k)
{
// Store the size of the string
int n = s.length();
// Store the total number of 1s
// and 0s in the entire string
int tot_ones = 0, tot_zeros = 0;
// Traverse the string S to find
// the frequency of 1 and 0
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == '1')
tot_ones++;
else
tot_zeros++;
}
// Store the number of 1s and
// 0s in the substring s[l, r]
int ones = 0, zeros = 0, sum = 0;
// Traverse the substring S[l, r]
// to find the frequency of 1s
// and 0s in it
for (int i = l - 1; i < r; i++) {
if (s.charAt(i) == '1') {
ones++;
sum++;
}
else
zeros++;
}
// Store the count of 1s and
// 0s outside substring s[l, r]
int rem_ones = tot_ones - ones;
int rem_zeros = tot_zeros - zeros;
// Check if the sum of the
// substring is at most K
if (k >= sum) {
// Store number of 1s required
int rem = k - sum;
// Check if there are enough 1s
// remaining to be swapped
if (zeros >= rem && rem_ones >= rem)
return rem;
}
// If the count of 1s in the substring exceeds k
else if (k < sum) {
// Store the number of 0s required
int rem = sum - k;
// Check if there are enough 0s
// remaining to be swapped
if (ones >= rem && rem_zeros >= rem)
return rem;
}
// In all other cases, print -1
return -1;
}
// Driver Code
public static void main(String[] args)
{
String S = "110011111000101";
int L = 5, R = 8, K = 2;
System.out.println(
minimumSwaps(S, L, R, K));
}
}
# Python3 program for the above approach
# Function to find the minimum number
# of swaps required such that the
# substring {s[l], .., s[r]} consists
# of exactly k 1s
def minimumSwaps(s, l, r, k) :
# Store the size of the string
n = len(s)
# Store the total number of 1s
# and 0s in the entire string
tot_ones, tot_zeros = 0, 0
# Traverse the string S to find
# the frequency of 1 and 0
for i in range(0, len(s)) :
if (s[i] == '1') :
tot_ones += 1
else :
tot_zeros += 1
# Store the number of 1s and
# 0s in the substring s[l, r]
ones, zeros, Sum = 0, 0, 0
# Traverse the substring S[l, r]
# to find the frequency of 1s
# and 0s in it
for i in range(l - 1, r) :
if (s[i] == '1') :
ones += 1
Sum += 1
else :
zeros += 1
# Store the count of 1s and
# 0s outside substring s[l, r]
rem_ones = tot_ones - ones
rem_zeros = tot_zeros - zeros
# Check if the sum of the
# substring is at most K
if (k >= Sum) :
# Store number of 1s required
rem = k - Sum
# Check if there are enough 1s
# remaining to be swapped
if (zeros >= rem and rem_ones >= rem) :
return rem
# If the count of 1s in the substring exceeds k
elif (k < Sum) :
# Store the number of 0s required
rem = Sum - k
# Check if there are enough 0s
# remaining to be swapped
if (ones >= rem and rem_zeros >= rem) :
return rem
# In all other cases, print -1
return -1
S = "110011111000101"
L, R, K = 5, 8, 2
print(minimumSwaps(S, L, R, K))
# This code is contributed by divyeshrabadiya07.
// C# program to implement
// the above approach
using System;
class GFG
{
// Function to find the minimum number
// of swaps required such that the
// substring {s[l], .., s[r]} consists
// of exactly k 1s
static int minimumSwaps(string s, int l, int r, int k)
{
// Store the size of the string
int n = s.Length;
// Store the total number of 1s
// and 0s in the entire string
int tot_ones = 0, tot_zeros = 0;
// Traverse the string S to find
// the frequency of 1 and 0
for (int i = 0; i < s.Length; i++)
{
if (s[i] == '1')
tot_ones++;
else
tot_zeros++;
}
// Store the number of 1s and
// 0s in the substring s[l, r]
int ones = 0, zeros = 0, sum = 0;
// Traverse the substring S[l, r]
// to find the frequency of 1s
// and 0s in it
for (int i = l - 1; i < r; i++)
{
if (s[i] == '1')
{
ones++;
sum++;
}
else
zeros++;
}
// Store the count of 1s and
// 0s outside substring s[l, r]
int rem_ones = tot_ones - ones;
int rem_zeros = tot_zeros - zeros;
// Check if the sum of the
// substring is at most K
if (k >= sum)
{
// Store number of 1s required
int rem = k - sum;
// Check if there are enough 1s
// remaining to be swapped
if (zeros >= rem && rem_ones >= rem)
return rem;
}
// If the count of 1s in the substring exceeds k
else if (k < sum)
{
// Store the number of 0s required
int rem = sum - k;
// Check if there are enough 0s
// remaining to be swapped
if (ones >= rem && rem_zeros >= rem)
return rem;
}
// In all other cases, print -1
return -1;
}
// Driver Code
public static void Main(String[] args)
{
string S = "110011111000101";
int L = 5, R = 8, K = 2;
Console.Write(minimumSwaps(S, L, R, K));
}
}
// This code is contributed by splevel62.
<script>
// javascript program for the above approach
// Function to find the minimum number
// of swaps required such that the
// substring {s[l], .., s[r]} consists
// of exactly k 1s
function minimumSwaps(s , l , r , k)
{
// Store the size of the string
var n = s.length;
// Store the total number of 1s
// and 0s in the entire string
var tot_ones = 0, tot_zeros = 0;
// Traverse the string S to find
// the frequency of 1 and 0
for (i = 0; i < s.length; i++) {
if (s.charAt(i) == '1')
tot_ones++;
else
tot_zeros++;
}
// Store the number of 1s and
// 0s in the substring s[l, r]
var ones = 0, zeros = 0, sum = 0;
// Traverse the substring S[l, r]
// to find the frequency of 1s
// and 0s in it
for (var i = l - 1; i < r; i++) {
if (s.charAt(i) == '1') {
ones++;
sum++;
}
else
zeros++;
}
// Store the count of 1s and
// 0s outside substring s[l, r]
var rem_ones = tot_ones - ones;
var rem_zeros = tot_zeros - zeros;
// Check if the sum of the
// substring is at most K
if (k >= sum) {
// Store number of 1s required
var rem = k - sum;
// Check if there are enough 1s
// remaining to be swapped
if (zeros >= rem && rem_ones >= rem)
return rem;
}
// If the count of 1s in the substring exceeds k
else if (k < sum) {
// Store the number of 0s required
var rem = sum - k;
// Check if there are enough 0s
// remaining to be swapped
if (ones >= rem && rem_zeros >= rem)
return rem;
}
// In all other cases, print -1
return -1;
}
// Driver Code
var S = "110011111000101";
var L = 5, R = 8, K = 2;
document.write(
minimumSwaps(S, L, R, K));
// This code is contributed by 29AjayKumar
</script>
Time Complexity: O(N)
Auxiliary Space: O(1)
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