XOR(^) is a binary operation called exclusive OR and works as

1^0 = 1  
0^1 = 1  
0^0 = 0
1^1 = 0  

XOR by 1 can work like a toggle switch that turns 1 to 0 or 0 to 1.

Another interesting thing to note is

x^0 = x 
x^x = 0

Usage:

Problem 1: Given a number . Flip all bits in its binary representation.

Solution 1: ^ considering is bit integer.

Problem 2: Given two numbers and . Swap and without using airthmetic operator and without using third variable.

Solution 2:

= ^

= ^

= ^

This is a simple operation. Suppose we need the index of the maximum element.

max_val = -1 \\initialise max
index = -1 \\take some initial impossible index to consider possibility of element not found.  
for (int i = 0;i<length;i++) {
    if (arr[i]>max_val) {
        max_val = arr[i];
        index = i;
    }
}

Similarly, we can get minimum value and index.

We can also collect multiple maximum values by making a small modification as

\\let container be a vector or a list 
if (arr[i]>max_val) {
    max_val = arr[i];
    index = i;
    container.clear();
}
else if (arr[i] == max_val) {
    container.push_back(i);
}