Given an integer N, task is to find the numbers which when raised to the power of 2 and added finally, gives the integer N.
Input : 71307 Output : 0, 1, 3, 7, 9, 10, 12, 16 Explanation : 71307 = 2^0 + 2^1 + 2^3 + 2^7 + 2^9 + 2^10 + 2^12 + 2^16 Input : 1213 Output : 0, 2, 3, 4, 5, 7, 10 Explanation : 1213 = 2^0 + 2^2 + 2^3 + 2^4 + 2^5 + 2^7 + 2^10
Every number can be described in powers of 2.
Example : 29 = 2^0 + 2^2 + 2^3 + 2^4.
2^0 ( exponent of 2 is ‘0’) 0
2^2 ( exponent of 2 is ‘2’) 1
2^3 ( exponent of 2 is ‘3’) 3
2^4 ( exponent of 2 is ‘4’) 4
Convert each number into its binary equivalent by pushing remainder of given number, when divided by 2 till it is greater than 0, to vector. Now, Iterate through its binary equivalent and whenever there is set bit, just print the i-th value(iteration number).
Hamming Code : Hamming Code is an error correcting code which can detect and correct one bit error. This pattern is also used in Hamming code error detection where parity bits store the XOR of numbers on the basis of LSB(Least Significant bit), where numbers are assigned in blocks and you need to find the blocks where the sum of power of 2 resulting to given number exists. Below is the image to show the blocks with given numbers.
Below is the implementation of above approach :
Blocks for 71307 : 0, 1, 3, 7, 9, 10, 12, 16 Blocks for 1213 : 0, 2, 3, 4, 5, 7, 10 Blocks for 29 : 0, 2, 3, 4 Blocks for 100 : 2, 5, 6
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