http://www.slideshare.net/gkumar007/bits-next-higher-presentation
5.3 Bits – next smallest and largest have the same No. of 1′s. | Ruobo [m2w]
5.3 Given an integer, print the next smallest and next largest number that have the same number of 1 bits in their binary representation.
The Brute Force Approach:
An easy approach is simply brute force: count the number of 1’s in n, and then increment (or decrement) until you find a number with the same number of 1’s. Easy
Observations:
»»If we “turn on” a 0, we need to “turn off” a 1
»»If we turn on a 0 at bit i and turn off a 1 at bit j, the number changes by 2^i – 2^j.
»»If we want to get a bigger number with the same number of 1s and 0s, i must be bigger than j.
Solution:
Code from http://ms-amazon.blogspot.com/2013/06/given-number-print-nex-largest-number.html
Also refer to http://www.slideshare.net/gkumar007/bits-next-higher-presentation
Read full article from 5.3 Bits – next smallest and largest have the same No. of 1′s. | Ruobo [m2w]
5.3 Bits – next smallest and largest have the same No. of 1′s. | Ruobo [m2w]
5.3 Given an integer, print the next smallest and next largest number that have the same number of 1 bits in their binary representation.
The Brute Force Approach:
An easy approach is simply brute force: count the number of 1’s in n, and then increment (or decrement) until you find a number with the same number of 1’s. Easy
Observations:
»»If we “turn on” a 0, we need to “turn off” a 1
»»If we turn on a 0 at bit i and turn off a 1 at bit j, the number changes by 2^i – 2^j.
»»If we want to get a bigger number with the same number of 1s and 0s, i must be bigger than j.
Solution:
next smallest
1. Traverse from right to left. Once we’ve passed a 1, turn on the next 0. We’ve now increased the number by 2^i. Yikes! Example: xxxxx011100 becomes xxxxx111100
2. Turn off the one that’s just to the right side of that. We’re now bigger by 2^i – 2^(i-1) Example: xxxxx111100 becomes xxxxx101100
3. Make the number as small as possible by rearranging all the 1s to be as far right as possible: Example: xxxxx101100 becomes xxxxx100011
1. Traverse from right to left. Once we’ve passed a 1, turn on the next 0. We’ve now increased the number by 2^i. Yikes! Example: xxxxx011100 becomes xxxxx111100
2. Turn off the one that’s just to the right side of that. We’re now bigger by 2^i – 2^(i-1) Example: xxxxx111100 becomes xxxxx101100
3. Make the number as small as possible by rearranging all the 1s to be as far right as possible: Example: xxxxx101100 becomes xxxxx100011
previous largest, we do the reverse.
1. Traverse from right to left. Once we’ve passed a zero, turn off the next 1. Example: xxxxx100011 becomes xxxxx000011.
2. Turn on the 0 that is directly to the right. Example: xxxxx000011 becomes xxxxx010011.
3. Make the number as big as possible by shifting all the ones as far to the left as possible. Example: xxxxx010011 becomes xxxxx011100 .
1. Traverse from right to left. Once we’ve passed a zero, turn off the next 1. Example: xxxxx100011 becomes xxxxx000011.
2. Turn on the 0 that is directly to the right. Example: xxxxx000011 becomes xxxxx010011.
3. Make the number as big as possible by shifting all the ones as far to the left as possible. Example: xxxxx010011 becomes xxxxx011100 .
next smallest
xxxxx011100 ↓ find a 1, make next 0=1,
xxxxx101100 ↓ make last 1=0,
xxxxx100011 push all right 1's to right.
previous largest, we do the reverse.
xxxxx100011 ↓ find a 0, make next 1=0, and set its preivous 0=1
xxxxx010011 ↓ then push every 1's to the left.
xxxxx011100
//check if bit at pos is 1 or 0 bool isSet(int number , int pos) { if( (number & (1<<pos)) > 0) return true; return false; } //set Bit at index pos to 1 int setBit(int number , int pos) { number |= 1<<pos; return number; } //set bit at index pos to 0 int unsetBit(int number, int pos) { number &= ~(1<<pos); return number; } //Return Next Greater number int getNextGreater(int number) { if(number <= 0) return -1; int maxIndex; int countOnes = 0; int pos = 0; //Scan number from right to left bit. for(bool encounterFlag = false; pos < 32 ; pos++) { if(encounterFlag) { if( !isSet(number , pos) ) { // set first 0 bit after encounteredFlag is true number = setBit(number, pos); // unset 1 bit immediately right of the above bit. number = unsetBit(number, --pos); break; } else continue; } //As soon as a 1 is encountered, encounteredFlag is set. if( isSet(number , pos)) encounterFlag = true; } //Count no. of 1's after maxIndex. for(int i=pos -1; i>=0 ; i--) { if(isSet(number, i)) countOnes++; } //Pushing all 1's to left. for(int j=pos -1; j>= countOnes; j--) number = unsetBit(number, j); for(int k=countOnes-1 ; k>=0; k--) number = setBit(number, k); return number; } int getNextSmaller(int number) { if(number < 0) return -1; int maxIndex; int countOnes = 0; int pos = 0; for(bool encounterFlag = false; pos < 32; pos++) { if(encounterFlag) { if(isSet(number, pos)) { //After encounterFlag is set Turnoff next 1 Bit. number = unsetBit(number, pos); //Turn on 0 bit on right of this bit. number = setBit(number, --pos); break; } else continue; } //Set encounter flag after first zero is encountered. if(!isSet(number,pos)) { encounterFlag = true; } } //Count no. of 1's after maxIndex. for(int i=pos -1; i>=0 ; i--) { if(isSet(number, i)) countOnes++; } //Pushing all 1's to left. for(int j=pos -1; j>= countOnes; j--) number = setBit(number, j); for(int k=countOnes-1 ; k>=0; k--) number = unsetBit(number, k); return number; } int main() { cout<<getNextGreater(18)<<endl; cout<<getNextSmaller(18)<<endl; system("pause"); }Better SolutionO(1) for Next higher number with same number of set bits http://www.geeksforgeeks.org/next-higher-number-with-same-number-of-set-bits/
Also refer to http://www.slideshare.net/gkumar007/bits-next-higher-presentation
Read full article from 5.3 Bits – next smallest and largest have the same No. of 1′s. | Ruobo [m2w]