# Minimum splits in a binary string such that every substring is a power of 4 or 6.

• Difficulty Level : Hard
• Last Updated : 23 Apr, 2021

Given a string S composed of 0 and 1. Find the minimum splits such that the substring is a binary representation of the power of 4 or 6 with no leading zeros. Print -1 if no such partitioning is possible.
Examples:

```Input: 100110110
Output: 3
The string can be split into a minimum of
three substrings 100(power of 4), 110
(power of 6) and 110(power of 6).

Input : 00000
Output : -1
0 is not a power of  4 or 6.```

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A simple solution is to split the string recursively at different indices and check if each split is a power of 4 or 6. Start with index 0 and split str[0] from other string. If it is a power of 4 or 6 then call recursively for index 1 and perform the same operation. When an entire string is split check if a total number of partitions are minimum so far or not. Then split str[0..1], check if it is the power of 4 or 6 and then call recursively for rest string. Compare partitions with minimum so far at the end of string traversal. This approach will be exponential in time.
An efficient solution is to use Dynamic Programming. A 1-D dp table is created in which dp[i] stores minimum number of partitions required to split string str[i..n-1] into substrings that are power of 4 or 6. Suppose we are at index i and str[i..j] is power of 4 or 6, then minimum number of partitions will be minimum number of partitions to split str[j+1..n-1] plus one partition to split str[i..j] from string, that is dp[j+1] + 1. Hence the recurrence relation for (j!=(n-1)) and (dp[j + 1]!=-1) will be:

dp[i] = min(dp[i], dp[j + 1] + 1)

Implementation:

## C++

 `// CPP program for Minimum splits in a``//string such that substring is a power of 4 or 6.` `#include ``using` `namespace` `std;` `// Function to find if given number``// is power of another number or not.``bool` `isPowerOf(``long` `val, ``int` `base)``{` `    ``// Divide given number repeatedly``    ``// by base value.``    ``while` `(val > 1) {``        ``if` `(val % base != 0)``            ``return` `false``; ``// not a power``        ``val /= base;``    ``}` `    ``return` `true``;``}` `// Function to find minimum number of``// partitions of given binary string``// so that each partition is power of 4 or 6.``int` `numberOfPartitions(string binaryNo)``{``    ``int` `i, j, n = binaryNo.length();` `    ``// Variable to store integer value of``    ``// given binary string partition.``    ``long` `val;` `    ``// DP table to store results of``    ``// partitioning done at differentindices.``    ``int` `dp[n];` `    ``// If the last digit is 1, hence 4^0=1 and 6^0=1``    ``dp[n - 1] = ((binaryNo[n - 1] - ``'0'``) == 0) ? -1 : 1;` `    ``// Fix starting position for partition``    ``for` `(i = n - 2; i >= 0; i--) {``        ``val = 0;` `        ``// Binary representation``        ``// with leading zeroes is not allowed.``        ``if` `((binaryNo[i] - ``'0'``) == 0) {``            ``dp[i] = -1;``            ``continue``;``        ``}` `        ``dp[i] = INT_MAX;` `        ``// Iterate for all different partitions starting from i``        ``for` `(j = i; j < n; j++) {` `            ``// Find integer value of current``            ``// binary partition.``            ``val = (val * 2) + (``long``)(binaryNo[j] - ``'0'``);` `            ``// Check if the value is a power of 4 or 6 or not``            ``// apply recurrence relation``            ``if` `(isPowerOf(val, 4) || isPowerOf(val, 6)) {``                ``if` `(j == n - 1) {``                    ``dp[i] = 1;``                ``}``                ``else` `{``                    ``if` `(dp[j + 1] != -1)``                        ``dp[i] = min(dp[i], dp[j + 1] + 1);``                ``}``            ``}``        ``}` `        ``// If no partitions are possible, then``        ``// make dp[i] = -1 to represent this.``        ``if` `(dp[i] == INT_MAX)``            ``dp[i] = -1;``    ``}` `    ``return` `dp[0];``}` `// Driver code``int` `main()``{``    ``string binaryNo = ``"100110110"``;``    ``cout << numberOfPartitions(binaryNo);``    ``return` `0;``}`

## Java

 `// Java program for Minimum splits``// in a string such that substring``// is a power of 4 or 6.``import` `java.io.*;` `class` `GFG``{``    ``static` `boolean` `isPowerOf(``long` `val,``                             ``int` `base)``{` `    ``// Divide given number``    ``// repeatedly by base value.``    ``while` `(val > ``1``)``    ``{``        ``if` `(val % base != ``0``)``            ``return` `false``; ``// not a power``        ``val /= base;``    ``}` `    ``return` `true``;``}` `// Function to find minimum``// number of partitions of``// given binary string so that``// each partition is power``// of 4 or 6.``static` `int` `numberOfPartitions(String binaryNo)``{``    ``int` `i, j, n = binaryNo.length();` `    ``// Variable to store integer``    ``// value of given binary``    ``// string partition.``    ``long` `val;` `    ``// DP table to store results``    ``// of partitioning done at``    ``// differentindices.``    ``int` `dp[] = ``new` `int``[n];` `    ``// If the last digit is 1,``    ``// hence 4^0=1 and 6^0=1``    ``dp[n - ``1``] = (((binaryNo.charAt(n - ``1``) -``                               ``'0'``) == ``0``) ?``                                   ``-``1` `: ``1``);` `    ``// Fix starting position``    ``// for partition``    ``for` `(i = n - ``2``; i >= ``0``; i--)``    ``{``        ``val = ``0``;` `        ``// Binary representation``        ``// with leading zeroes``        ``// is not allowed.``        ``if` `((binaryNo.charAt(i) - ``'0'``) == ``0``)``        ``{``            ``dp[i] = -``1``;``            ``continue``;``        ``}` `        ``dp[i] = Integer.MAX_VALUE;` `        ``// Iterate for all different``        ``// partitions starting from i``        ``for` `(j = i; j < n; j++)``        ``{` `            ``// Find integer value of``            ``// current binary partition.``            ``val = (val * ``2``) +``                  ``(``long``)(binaryNo.charAt(j) - ``'0'``);` `            ``// Check if the value is a``            ``// power of 4 or 6 or not``            ``// apply recurrence relation``            ``if` `(isPowerOf(val, ``4``) ||``                ``isPowerOf(val, ``6``))``            ``{``                ``if` `(j == n - ``1``)``                ``{``                    ``dp[i] = ``1``;``                ``}``                ``else``                ``{``                    ``if` `(dp[j + ``1``] != -``1``)``                        ``dp[i] = Math.min(dp[i],``                                         ``dp[j + ``1``] + ``1``);``                ``}``            ``}``        ``}` `        ``// If no partitions are possible,``        ``// then make dp[i] = -1 to``        ``// represent this.``        ``if` `(dp[i] == Integer.MAX_VALUE)``            ``dp[i] = -``1``;``    ``}` `    ``return` `dp[``0``];``}` `// Driver code``public` `static` `void` `main (String[] args)``{``    ``String binaryNo = ``"100110110"``;``    ``System.out.println(numberOfPartitions(binaryNo));``}``}` `// This code is contributed``// by shiv_bhakt.`

## Python 3

 `# Python 3 program for Minimum``# splits in a string such that``# substring is a power of 4 or 6.` `import` `sys` `# Function to find if given number``# is power of another number or not.``def` `isPowerOf(val, base):` `    ``# Divide given number repeatedly``    ``# by base value.``    ``while` `(val > ``1``):``        ``if` `(val ``%` `base !``=` `0``):``            ``return` `False` `# not a power``        ``val ``/``/``=` `base` `    ``return` `True` `# Function to find minimum number of``# partitions of given binary string``# so that each partition is power of 4 or 6.``def` `numberOfPartitions(binaryNo):` `    ``n ``=` `len``(binaryNo)` `    ``# DP table to store results of``    ``# partitioning done at differentindices.``    ``dp ``=` `[``0``] ``*` `n` `    ``# If the last digit is 1, hence 4^0=1 and 6^0=1``    ``if` `((``ord``(binaryNo[n ``-` `1``]) ``-` `ord``(``'0'``)) ``=``=` `0``) :``        ``dp[n ``-` `1``] ``=` `-``1``    ``else``:``        ``dp[n ``-` `1``] ``=` `1` `    ``# Fix starting position for partition``    ``for` `i ``in` `range``( n ``-` `2``, ``-``1``, ``-``1``):``        ``val ``=` `0` `        ``# Binary representation``        ``# with leading zeroes is not allowed.``        ``if` `((``ord``(binaryNo[i]) ``-` `ord``(``'0'``)) ``=``=` `0``):``            ``dp[i] ``=` `-``1``            ``continue` `        ``dp[i] ``=` `sys.maxsize` `        ``# Iterate for all different partitions starting from i``        ``for` `j ``in` `range``(i, n):` `            ``# Find integer value of current``            ``# binary partition.``            ``val ``=` `(val ``*` `2``) ``+` `(``ord``(binaryNo[j]) ``-` `ord``(``'0'``))` `            ``# Check if the value is a power of 4 or 6 or not``            ``# apply recurrence relation``            ``if` `(isPowerOf(val, ``4``) ``or` `isPowerOf(val, ``6``)):``                ``if` `(j ``=``=` `n ``-` `1``):``                    ``dp[i] ``=` `1``                ` `                ``else` `:``                    ``if` `(dp[j ``+` `1``] !``=` `-``1``):``                        ``dp[i] ``=` `min``(dp[i], dp[j ``+` `1``] ``+` `1``)` `        ``# If no partitions are possible, then``        ``# make dp[i] = -1 to represent this.``        ``if` `(dp[i] ``=``=` `sys.maxsize):``            ``dp[i] ``=` `-``1` `    ``return` `dp[``0``]` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ` `    ``binaryNo ``=` `"100110110"``    ``print``(numberOfPartitions(binaryNo))``    ` `# This code is contributed by Ita_c.   `

## C#

 `// C# program for Minimum splits``// in a string such that substring``// is a power of 4 or 6.` `using` `System;`` ` `class` `GFG``{``    ``static` `bool` `isPowerOf(``long` `val, ``int` `b)``{`` ` `    ``// Divide given number``    ``// repeatedly by base value.``    ``while` `(val > 1)``    ``{``        ``if` `(val % b != 0)``            ``return` `false``; ``// not a power``        ``val /= b;``    ``}`` ` `    ``return` `true``;``}`` ` `// Function to find minimum``// number of partitions of``// given binary string so that``// each partition is power``// of 4 or 6.``static` `int` `numberOfPartitions(``string` `binaryNo)``{``    ``int` `i, j, n = binaryNo.Length;`` ` `    ``// Variable to store integer``    ``// value of given binary``    ``// string partition.``    ``long` `val;`` ` `    ``// DP table to store results``    ``// of partitioning done at``    ``// differentindices.``    ``int``[] dp = ``new` `int``[n];`` ` `    ``// If the last digit is 1,``    ``// hence 4^0=1 and 6^0=1``    ``dp[n - 1] = (((binaryNo[n - 1] -``                               ``'0'``) == 0) ?``                                   ``-1 : 1);`` ` `    ``// Fix starting position``    ``// for partition``    ``for` `(i = n - 2; i >= 0; i--)``    ``{``        ``val = 0;`` ` `        ``// Binary representation``        ``// with leading zeroes``        ``// is not allowed.``        ``if` `((binaryNo[i] - ``'0'``) == 0)``        ``{``            ``dp[i] = -1;``            ``continue``;``        ``}`` ` `        ``dp[i] = ``int``.MaxValue;`` ` `        ``// Iterate for all different``        ``// partitions starting from i``        ``for` `(j = i; j < n; j++)``        ``{`` ` `            ``// Find integer value of``            ``// current binary partition.``            ``val = (val * 2) +``                  ``(``long``)(binaryNo[j] - ``'0'``);`` ` `            ``// Check if the value is a``            ``// power of 4 or 6 or not``            ``// apply recurrence relation``            ``if` `(isPowerOf(val, 4) ||``                ``isPowerOf(val, 6))``            ``{``                ``if` `(j == n - 1)``                ``{``                    ``dp[i] = 1;``                ``}``                ``else``                ``{``                    ``if` `(dp[j + 1] != -1)``                        ``dp[i] = Math.Min(dp[i],``                                         ``dp[j + 1] + 1);``                ``}``            ``}``        ``}`` ` `        ``// If no partitions are possible,``        ``// then make dp[i] = -1 to``        ``// represent this.``        ``if` `(dp[i] == ``int``.MaxValue)``            ``dp[i] = -1;``    ``}`` ` `    ``return` `dp[0];``}`` ` `// Driver code``public` `static` `void` `Main ()``{``    ``string` `binaryNo = ``"100110110"``;``    ``Console.Write(numberOfPartitions(binaryNo));``}``}`

## Javascript

 ``

Output:

` 3`

Time Complexity: O(n^2*log(x)), x = largest power of 4 or 6 obtainable from input string.
Auxiliary Space: O(n)

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