Divide two integers without using division, multiplication and mod operator, Flaw in bitmanipulation loop breaking equation

Question

This question has been answered here.

My query is, following approach-1 works, however the variation of it, that is approach-2 does not, rather it gives time limit exceeded. I can not find out why. I used interviewbit portal to solve this question, and here it is in leetcode.

The issue is

for(; i<32 && ((B<<i)-A <=0); i++) {
}

works, but

for(; i<32 && (B<<i)<=A; i++) {
}

does not work, rather gives Time Limit Exceeds issue.

Approach-1

public class Solution {
    public int divide(int A, int B) {

        //Base Cases
        if (B == 1) {
            return A;
        }
        if (A == Integer.MIN_VALUE && B == -1) {
            return Integer.MAX_VALUE;
        }

        boolean neg = false;
        if((A<0 && B>0) || (A>0 && B<0)) neg=true;
        A = A==Integer.MIN_VALUE ? Integer.MAX_VALUE : Math.abs(A);
        B = Math.abs(B);
        if(A<B) return 0;

        int res=0;
        while(A>=B) {
            int i=0;

            for(; i<32 && ((B<<i)-A <=0); i++) {
            }

            i--;
            res+=(1<<i);
            A-=(B<<i);
        }
      return neg? res*-1 : res;
    }
}

Approach-2

public class Solution {
    public int divide(int A, int B) {

        //Base Cases
        if (B == 1) {
            return A;
        }
        if (A == Integer.MIN_VALUE && B == -1) {
            return Integer.MAX_VALUE;
        }

        boolean neg = false;
        if((A<0 && B>0) || (A>0 && B<0)) neg=true;
        A = A==Integer.MIN_VALUE ? Integer.MAX_VALUE : Math.abs(A);
        B=Math.abs(B);
        if(A<B) return 0;
        int res=0;

        while(A>=B) {
            int i=0;

            for(; i<32 && (B<<i)<=A; i++) {
            }

            i--;
            res+=(1<<i);
            A-=(B<<i);
        }
      return neg? res*-1 : res;
    }
}