Minimum number of operations required to sum to binary string S

Given a binary string, S. Find the minimum number of operations required to be performed on the number zero to convert it to the number represented by S.
It is allowed to perform operations of 2 types:

• Add 2^x
• Subtract 2^x

Note: Start performing operations on 0.

Examples:

Input : 100 
Output : 1 
Explanation: We just perform a single operation, i.e Add 4 (2^2)

Input : 111 
Output : 2 
Explanation: We perform the following two operations: 
1) Add 8(2^3) 
2) Subtract 1(2^0)

The idea is to use Dynamic Programming to solve this problem.
Note: In the below analysis, we considered that all binary strings are represented from LSB to MSB (from LHS to RHS) i.e 2(Binary form: “10”) is represented as “01”.
Let the given binary string be S.
Let dp[i][0] represent the minimum number of operations required to make a binary string R such that R[0…i] is the same as S[0…i] and R[i+1…] = “00..0”
Similarly, let dp[i][1] represent the minimum number of operations required to make a binary string R such that R[0…i] is the same as S[0…i] and R[i+1…] = “11..1”
If S_i is ‘0’, then dp[i][0] = dp[i – 1][0]. Since we do not require any additional operations. Now we consider the value of dp[i][1], which is a bit trickier. For dp[i][1], we can either make our transition from dp[i – 1][1] by just making the i^th character of the string formed by dp[i-1][1], ‘0’. Since earlier this, the i^th character was “1”. We just need to subtract 2ⁱ from the string represented by dp[i-1][0]. Thus, we perform one operation other than the ones represented by dp[i-1][0]. 
The other transition could be from dp[i-1][0]. Let dp[i-1][1] represent the string R. Then we need to keep R[i] = 0 as it already is but R[i + 1…..] which is currently “000..0”, needs to be changed to “111…1”, this can be done by subtracting 2⁽ⁱ⁺¹⁾ from R. Thus we just need one operation other than the ones represented by dp[i-1][0].
Similar is the case when S_i is ‘1’.
The final answer is the one represented by dp[l-1][0], where l is the length of the binary string S.
Below is the implementation of the above approach:

C++

// CPP program to find the minimum number
// of operations required to sum to N
#include <bits/stdc++.h>

using namespace std;

// Function to return the minimum operations required
// to sum to a number represented by the binary string S
int findMinOperations(string S)
{
    // Reverse the string to consider
    // it from LSB to MSB
    reverse(S.begin(), S.end());
    int n = S.length();

    // initialise the dp table
    int dp[n + 1][2];

    // If S[0] = '0', there is no need to
    // perform any operation
    if (S[0] == '0') {
        dp[0][0] = 0;
    }

    else {
        // If S[0] = '1', just perform a single
        // operation(i.e Add 2^0)
        dp[0][0] = 1;
    }

    // Irrespective of the LSB, dp[0][1] is always
    // 1 as there is always the need of making the
    // suffix of the binary string of the form "11....1"
    // as suggested by the definition of dp[i][1]
    dp[0][1] = 1;

    for (int i = 1; i < n; i++) {
        if (S[i] == '0') {

            // Transition from dp[i - 1][0]
            dp[i][0] = dp[i - 1][0];

            /* 1. Transition from dp[i - 1][1] by just doing
                  1 extra operation of subtracting 2^i
               2. Transition from dp[i - 1][0] by just doing
                  1 extra operation of subtracting 2^(i+1) */
            dp[i][1] = 1 + min(dp[i - 1][1], dp[i - 1][0]);
        }
        else {

            // Transition from dp[i - 1][1]
            dp[i][1] = dp[i - 1][1];

            /* 1. Transition from dp[i - 1][1] by just doing
                  1 extra operation of adding 2^(i+1)
               2. Transition from dp[i - 1][0] by just doing
                  1 extra operation of adding 2^i */
            dp[i][0] = 1 + min(dp[i - 1][0], dp[i - 1][1]);
        }
    }

    return dp[n - 1][0];
}

// Driver Code
int main()
{
    string S = "100";
    cout << findMinOperations(S) << endl;

    S = "111";
    cout << findMinOperations(S) << endl;

    return 0;
}

Java

// Java program to find the minimum number
// of operations required to sum to N

class GFG
{

    // Function to return the minimum operations required
    // to sum to a number represented by the binary string S
    static int findMinOperations(String S)
    {

        // Reverse the string to consider
        // it from LSB to MSB
        S = reverse(S);
        int n = S.length();

        // initialise the dp table
        int dp[][] = new int[n + 1][2];

        // If S[0] = '0', there is no need to
        // perform any operation
        if (S.charAt(0) == '0')
        {
            dp[0][0] = 0;
        }
        else
        {
            // If S[0] = '1', just perform a single
            // operation(i.e Add 2^0)
            dp[0][0] = 1;
        }

        // Irrespective of the LSB, dp[0][1] is always
        // 1 as there is always the need of making the
        // suffix of the binary string of the form "11....1"
        // as suggested by the definition of dp[i][1]
        dp[0][1] = 1;

        for (int i = 1; i < n; i++)
        {
            if (S.charAt(i) == '0')
            {
                // Transition from dp[i - 1][0]
                dp[i][0] = dp[i - 1][0];

                /* 1. Transition from dp[i - 1][1] by just doing
                1 extra operation of subtracting 2^i
                2. Transition from dp[i - 1][0] by just doing
                1 extra operation of subtracting 2^(i+1) */
                dp[i][1] = 1 + Math.min(dp[i - 1][1], dp[i - 1][0]);
            }
            else
            {

                // Transition from dp[i - 1][1]
                dp[i][1] = dp[i - 1][1];

                /* 1. Transition from dp[i - 1][1] by just doing
                1 extra operation of adding 2^(i+1)
                2. Transition from dp[i - 1][0] by just doing
                1 extra operation of adding 2^i */
                dp[i][0] = 1 + Math.min(dp[i - 1][0], dp[i - 1][1]);
            }
        }

        return dp[n - 1][0];
    }

    static String reverse(String input)
    {
        char[] temparray = input.toCharArray();
        int left, right = 0;
        right = temparray.length - 1;
        for (left = 0; left < right; left++, right--)
        {
            // Swap values of left and right
            char temp = temparray[left];
            temparray[left] = temparray[right];
            temparray[right] = temp;
        }
        return String.valueOf(temparray);
    }

    // Driver Code
    public static void main(String[] args)
    {
        String S = "100";
        System.out.println(findMinOperations(S));
        S = "111";
        System.out.println(findMinOperations(S));
    }
}

// This code is contributed by
// PrinciRaj1992

Python3

# Python3 program to find the minimum
# number of operations required to sum to N

# Function to return the minimum
# operations required to sum to a
# number represented by the binary string S
def findMinOperations(S) :

    # Reverse the string to consider
    # it from LSB to MSB
    S = S[: : -1]
    n = len(S)

    # initialise the dp table
    dp = [[0] * 2] * (n + 1)

    # If S[0] = '0', there is no need
    # to perform any operation
    if (S[0] == '0') :
        dp[0][0] = 0

    else :

        # If S[0] = '1', just perform a
        # single operation(i.e Add 2^0)
        dp[0][0] = 1

    # Irrespective of the LSB, dp[0][1] is
    # always 1 as there is always the need
    # of making the suffix of the binary
    # string of the form "11....1" as
    # suggested by the definition of dp[i][1]
    dp[0][1] = 1

    for i in range(1, n) :

        if (S[i] == '0') :

            # Transition from dp[i - 1][0]
            dp[i][0] = dp[i - 1][0]

            """
            1. Transition from dp[i - 1][1]
                by just doing 1 extra operation
                of subtracting 2^i
            2. Transition from dp[i - 1][0] by
                just doing 1 extra operation of
                subtracting 2^(i+1)
            """
            dp[i][1] = 1 + min(dp[i - 1][1],
                               dp[i - 1][0])

        else :

            # Transition from dp[i - 1][1]
            dp[i][1] = dp[i - 1][1];

            """
            1. Transition from dp[i - 1][1] by
                just doing 1 extra operation
                of adding 2^(i+1)
            2. Transition from dp[i - 1][0] by
                just doing 1 extra operation
                of adding 2^i
            """
            dp[i][0] = 1 + min(dp[i - 1][0],
                               dp[i - 1][1])

    return dp[n - 1][0]

# Driver Code
if __name__ == "__main__" :

    S = "100"
    print(findMinOperations(S))

    S = "111";
    print(findMinOperations(S))

# This code is contributed by Ryuga

C#

// C# program to find the minimum number
// of operations required to sum to N
using System;

class GFG
{

    // Function to return the minimum
    // operations required to sum
    // to a number represented by
    // the binary string S
    static int findMinOperations(String S)
    {

        // Reverse the string to consider
        // it from LSB to MSB
        S = reverse(S);
        int n = S.Length;

        // initialise the dp table
        int [,]dp = new int[n + 1, 2];

        // If S[0] = '0', there is no need to
        // perform any operation
        if (S[0] == '0')
        {
            dp[0, 0] = 0;
        }
        else
        {
            // If S[0] = '1', just perform a single
            // operation(i.e Add 2^0)
            dp[0, 0] = 1;
        }

        // Irrespective of the LSB, dp[0,1] is always
        // 1 as there is always the need of making the
        // suffix of the binary string of the form "11....1"
        // as suggested by the definition of dp[i,1]
        dp[0, 1] = 1;

        for (int i = 1; i < n; i++)
        {
            if (S[i] == '0')
            {
                // Transition from dp[i - 1,0]
                dp[i, 0] = dp[i - 1, 0];

                /* 1. Transition from dp[i - 1,1] by just doing
                1 extra operation of subtracting 2^i
                2. Transition from dp[i - 1,0] by just doing
                1 extra operation of subtracting 2^(i+1) */
                dp[i, 1] = 1 + Math.Min(dp[i - 1, 1], dp[i - 1, 0]);
            }
            else
            {

                // Transition from dp[i - 1,1]
                dp[i, 1] = dp[i - 1, 1];

                /* 1. Transition from dp[i - 1,1] by just doing
                1 extra operation of adding 2^(i+1)
                2. Transition from dp[i - 1,0] by just doing
                1 extra operation of adding 2^i */
                dp[i, 0] = 1 + Math.Min(dp[i - 1, 0], dp[i - 1, 1]);
            }
        }
        return dp[n - 1, 0];
    }

    static String reverse(String input)
    {
        char[] temparray = input.ToCharArray();
        int left, right = 0;
        right = temparray.Length - 1;
        for (left = 0; left < right; left++, right--)
        {
            // Swap values of left and right
            char temp = temparray[left];
            temparray[left] = temparray[right];
            temparray[right] = temp;
        }
        return String.Join("",temparray);
    }

    // Driver Code
    public static void Main()
    {
        String S = "100";
        Console.WriteLine(findMinOperations(S));
        S = "111";
        Console.WriteLine(findMinOperations(S));
    }
}

//This code is contributed by 29AjayKumar

PHP

<?php
// PHP program to find the minimum number
// of operations required to sum to N

// Function to return the minimum operations required
// to sum to a number represented by the binary string S
function findMinOperations($S)
{
    // Reverse the string to consider
    // it from LSB to MSB
    $p= strrev($S);
    $n = strlen($p);

    // initialise the dp table
    $dp = array_fill(0,$n + 1,array_fill(0,2,NULL));

    // If S[0] = '0', there is no need to
    // perform any operation
    if ($p[0] == '0') {
        $dp[0][0] = 0;
    }

    else {
        // If S[0] = '1', just perform a single
        // operation(i.e Add 2^0)
        $dp[0][0] = 1;
    }

    // Irrespective of the LSB, dp[0][1] is always
    // 1 as there is always the need of making the
    // suffix of the binary string of the form "11....1"
    // as suggested by the definition of dp[i][1]
    $dp[0][1] = 1;

    for ($i = 1; $i < $n; $i++) {
        if ($p[$i] == '0') {

            // Transition from dp[i - 1][0]
            $dp[$i][0] = $dp[$i - 1][0];

            /* 1. Transition from dp[i - 1][1] by just doing
                  1 extra operation of subtracting 2^i
               2. Transition from dp[i - 1][0] by just doing
                  1 extra operation of subtracting 2^(i+1) */
            $dp[$i][1] = 1 + min($dp[$i - 1][1], $dp[$i - 1][0]);
        }
        else {

            // Transition from dp[i - 1][1]
            $dp[$i][1] = $dp[$i - 1][1];

            /* 1. Transition from dp[i - 1][1] by just doing
                  1 extra operation of adding 2^(i+1)
               2. Transition from dp[i - 1][0] by just doing
                  1 extra operation of adding 2^i */
            $dp[$i][0] = 1 + min($dp[$i - 1][0], $dp[$i - 1][1]);
        }
    }

    return $dp[$n - 1][0];
}

// Driver Code

$S = "100";
echo findMinOperations($S)."\n";

$S = "111";
echo findMinOperations($S)."\n";

return 0;

// This code is contributed by Ita_c.
?>

Javascript

<script>
// Javascript program to find the minimum number
// of operations required to sum to N

    // Function to return the minimum operations required
    // to sum to a number represented by the binary string S
    function findMinOperations(S)
    {
        // Reverse the string to consider
        // it from LSB to MSB
        S = reverse(S);
        let n = S.length;

        // initialise the dp table
        let dp = new Array(n + 1);
        for(let i=0;i<dp.length;i++)
        {
            dp[i]=new Array(2);
            for(let j=0;j<2;j++)
            {
                dp[i][j]=0;
            }
        }

        // If S[0] = '0', there is no need to
        // perform any operation
        if (S[0] == '0')
        {
            dp[0][0] = 0;
        }
        else
        {
            // If S[0] = '1', just perform a single
            // operation(i.e Add 2^0)
            dp[0][0] = 1;
        }

        // Irrespective of the LSB, dp[0][1] is always
        // 1 as there is always the need of making the
        // suffix of the binary string of the form "11....1"
        // as suggested by the definition of dp[i][1]
        dp[0][1] = 1;

        for (let i = 1; i < n; i++)
        {
            if (S[i] == '0')
            {
                // Transition from dp[i - 1][0]
                dp[i][0] = dp[i - 1][0];

                /* 1. Transition from dp[i - 1][1] by just doing
                1 extra operation of subtracting 2^i
                2. Transition from dp[i - 1][0] by just doing
                1 extra operation of subtracting 2^(i+1) */
                dp[i][1] = 1 + Math.min(dp[i - 1][1], dp[i - 1][0]);
            }
            else
            {

                // Transition from dp[i - 1][1]
                dp[i][1] = dp[i - 1][1];

                /* 1. Transition from dp[i - 1][1] by just doing
                1 extra operation of adding 2^(i+1)
                2. Transition from dp[i - 1][0] by just doing
                1 extra operation of adding 2^i */
                dp[i][0] = 1 + Math.min(dp[i - 1][0], dp[i - 1][1]);
            }
        }

        return dp[n - 1][0];
    }

    function reverse(input)
    {
        let temparray = input.split("");
        let left, right = 0;
        right = temparray.length - 1;
        for (left = 0; left < right; left++, right--)
        {
            // Swap values of left and right
            let temp = temparray[left];
            temparray[left] = temparray[right];
            temparray[right] = temp;
        }
        return (temparray).join("");
    }

    // Driver Code
    let S = "100";
    document.write(findMinOperations(S)+"<br>");

    S = "111";
    document.write(findMinOperations(S)+"<br>");


// This code is contributed by avanitrachhadiya2155
</script>

Output:

1
2

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