Minimum number of operations required to sum to binary string S

• Last Updated : 14 Nov, 2021

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:

• Subtract 2x

Note: Start performing operations on 0.

Examples:

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

Input : 111
Output :
Explanation: We perform the following two operations:
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] 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] 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 Si is ‘0’, then dp[i] = dp[i – 1]. Since we do not require any additional operations. Now we consider the value of dp[i], which is a bit trickier. For dp[i], we can either make our transition from dp[i – 1] by just making the ith character of the string formed by dp[i-1], ‘0’. Since earlier this, the ith character was “1”. We just need to subtract 2i from the string represented by dp[i-1]. Thus, we perform one operation other than the ones represented by dp[i-1].
The other transition could be from dp[i-1]. Let dp[i-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].
Similar is the case when Si is ‘1’.
The final answer is the one represented by dp[l-1], 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  using namespace std; // Function to return the minimum operations required// to sum to a number represented by the binary string Sint 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];     // If S = '0', there is no need to    // perform any operation    if (S == '0') {        dp = 0;    }     else {        // If S = '1', just perform a single        // operation(i.e Add 2^0)        dp = 1;    }     // Irrespective of the LSB, dp 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]    dp = 1;     for (int i = 1; i < n; i++) {        if (S[i] == '0') {             // Transition from dp[i - 1]            dp[i] = dp[i - 1];             /* 1. Transition from dp[i - 1] by just doing                  1 extra operation of subtracting 2^i               2. Transition from dp[i - 1] by just doing                  1 extra operation of subtracting 2^(i+1) */            dp[i] = 1 + min(dp[i - 1], dp[i - 1]);        }        else {             // Transition from dp[i - 1]            dp[i] = dp[i - 1];             /* 1. Transition from dp[i - 1] by just doing                  1 extra operation of adding 2^(i+1)               2. Transition from dp[i - 1] by just doing                  1 extra operation of adding 2^i */            dp[i] = 1 + min(dp[i - 1], dp[i - 1]);        }    }     return dp[n - 1];} // Driver Codeint 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];         // If S = '0', there is no need to        // perform any operation        if (S.charAt(0) == '0')        {            dp = 0;        }        else        {            // If S = '1', just perform a single            // operation(i.e Add 2^0)            dp = 1;        }         // Irrespective of the LSB, dp 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]        dp = 1;         for (int i = 1; i < n; i++)        {            if (S.charAt(i) == '0')            {                // Transition from dp[i - 1]                dp[i] = dp[i - 1];                 /* 1. Transition from dp[i - 1] by just doing                1 extra operation of subtracting 2^i                2. Transition from dp[i - 1] by just doing                1 extra operation of subtracting 2^(i+1) */                dp[i] = 1 + Math.min(dp[i - 1], dp[i - 1]);            }            else            {                 // Transition from dp[i - 1]                dp[i] = dp[i - 1];                 /* 1. Transition from dp[i - 1] by just doing                1 extra operation of adding 2^(i+1)                2. Transition from dp[i - 1] by just doing                1 extra operation of adding 2^i */                dp[i] = 1 + Math.min(dp[i - 1], dp[i - 1]);            }        }         return dp[n - 1];    }     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 Sdef findMinOperations(S) :     # Reverse the string to consider    # it from LSB to MSB    S = S[: : -1]    n = len(S)     # initialise the dp table    dp = [ * 2] * (n + 1)     # If S = '0', there is no need    # to perform any operation    if (S == '0') :        dp = 0         else :                 # If S = '1', just perform a        # single operation(i.e Add 2^0)        dp = 1         # Irrespective of the LSB, dp 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]    dp = 1     for i in range(1, n) :                 if (S[i] == '0') :             # Transition from dp[i - 1]            dp[i] = dp[i - 1]             """            1. Transition from dp[i - 1]                by just doing 1 extra operation                of subtracting 2^i            2. Transition from dp[i - 1] by                just doing 1 extra operation of                subtracting 2^(i+1)            """            dp[i] = 1 + min(dp[i - 1],                               dp[i - 1])                 else :             # Transition from dp[i - 1]            dp[i] = dp[i - 1];             """            1. Transition from dp[i - 1] by                just doing 1 extra operation                of adding 2^(i+1)            2. Transition from dp[i - 1] by                just doing 1 extra operation                of adding 2^i            """            dp[i] = 1 + min(dp[i - 1],                               dp[i - 1])     return dp[n - 1] # Driver Codeif __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 Nusing 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', there is no need to        // perform any operation        if (S == '0')        {            dp[0, 0] = 0;        }        else        {            // If S = '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



Javascript


Output:
1
2

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

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