C++ O(1) solution:

int maximizingXor(int l, int r) {
    int x = l ^ r, p = 1;
    while ((x & (p - 1)) != x) p <<= 1;
    return p - 1;
}

Why this works.. this looks for highest bit which is different between l and r: for example (11, 12) that is third bit with value 4.. all bits above it are identical in both l and r, and you can not produce any l<=a<=b<=r which have non zero result of a^b. But you can find value a which has third bit zero and all remaining bits set (11 in this example), and b = a + 1 will clear all lower bits and set third bit, so a^b is all ones from highest difference bit down to bottom bit = maximum xor difference. qed.

JAVA SOLUTION || Brute+BitManipulation

public static int maximizingXor(int l, int r) { // Write your code here int res=0; int xor=0; int fval=0; for(int i=l;i<=r;i++){ for(int j=i;j<=r;j++){ res=i^j; if(xor fval=xor; } return fval; }

JavaScript

  let max_value = 0;
    for(let i = l; i <= r; i++) {
        for(let j = i; j <= r; j++) {
            const v = i ^ j;
            if(max_value < v) {
                max_value = v;
            }
        }
    }
    return max_value;

Brute force (java):

    public static int maximizingXor(int l, int r) {
        int max = 0;
        
        for(int i = l; i < r; i++) {
            int xored = i ^ (i + 1);
            
            if(xored > max) {
                max = xored;
            }
        }
        
        return max;
    }

Java : T.C= O(r-l), Using properties of Bitwise operation. All-Correct

public static int maximizingXor(int l, int r) {

int max=Integer.MIN_VALUE;


int a,b;

a=l;

for(int i=l;i<r;i++)
 {
b=i+1;

if(a!=b)
    {
        if((a^b) > max)
        max=a^b;
     }
// assign a=b

b=a^b;
a=a^b;
}
        return max;
        }

}