## Interview Question

## Interview Answer

2 Answers

#include <string>

#include <cmath>

using namespace std;

string findBinSum(string a, string b) {

/*

Iterate through strings and find integer values.

Find sum. Convert sum back to binary string.

*/

int sizeA = a.size();

int sizeB = b.size();

int valA =0;

int valB =0;

for (int i=sizeA-1; i >= 0;i--)

{

if (a[i] == '1') {

valA += (int)pow((double)2,(double)(sizeA-1-i));

}

}

for (int k=sizeB-1; k >= 0;k--)

{

if (b[k] == '1') {

valB += (int)pow((double)2,(double)(sizeB-1-k));

}

}

cout << "valA = " << valA << endl;

cout << "valB = " << valB << endl;

int sum = valA + valB;

if (sum == 0) {string answer = "0"; return answer; }

cout << "the sum is " << sum << endl;

int power = (int)(log(sum)/log(2));

cout << "the power is " << power << endl;

int numUsed = (int)pow((double)2,(double)power);

sum -= numUsed;

string answer = "1";

for(int j=power-1; j>=0; j--) {

numUsed = (int)pow((double)2,(double)j);

if ( sum-numUsed >= 0) {

sum -= numUsed;

answer += "1";

}

else {

answer += "0";

}

}

return answer;

}

int main(){

string binary1 = "1011";

string binary2 = "1011";

string answer = findBinSum(binary1,binary2);

cout << "sum: " << answer << endl;

}

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// let A[1..n] n B[1..m] be the two strings and the result is to be stored in S[]

if(n>m)

k=n;

else

k=m;

C=0;

for(i=n, j=m; i>0 !! j>0; i--, j--)

{

if(i<=0) { i=0; A[0]=0}

if(j<=0) { j=0; B[0]=0}

count=0;

if(A[i]==1)

count++;

if(B[j]==1)

count++;

if(C==1)

count++;

switch(count)

{ case 0 : S[k]=0; C=0;

break;

case 1 : S[k]=1; C=0;

break;

case 2 : S[k]=0; C=1;

break;

case 3 : S[k]=1; C=1;

break;

}

k--;

}