These might be trivial for most users, but if you are like me, who sometimes confuse on when to use mid+1, rig-1 etc, here is a reliable method after considering many different implementation variants.
To find the index of the rightmost $$$1$$$ in a monotonically decreasing function, for example $$$a=[1,1,1,1,0,0,0]$$$
Define check(i) to return true when $$$a[i]=1$$$, false otherwise.
int lef = 0, rig = n;
while(lef < rig) {
int mid=(lef + rig)/2;
if(check(mid)){
lef = mid+1;
} else {
rig = mid;
}
}
int ans = lef-1;
Notice that the mid in lef=mid+1 is always valid, so the answer (lef-1) is the last valid mid. Also notice the initial values of lef and rig. rig is set out of bounds. If $$$a$$$ is available or can be computed efficiently, you can instead do
F0R(i,n)a[i]*=-1;
int ans = upper_bound(a,a+n,-1)-a-1; // get index of rightmost 1
Careful if $$$1$$$ doesn't exist in $$$a$$$.
To find the index of the leftmost $$$1$$$ in a monotonically increasing function, for example $$$a=[0,0,0,0,1,1,1]$$$
In a similar way,
int lef = -1, rig = n-1;
while(lef < rig) {
int mid=(lef + rig + 1)/2;
if(check(mid)){
rig = mid-1;
} else {
lef = mid;
}
}
int ans = rig+1;
Notice the ceiling in the definition of mid. This is to handle rig-lef equals one. Again notice the initial values for lef and rig. lef is set out of bounds. If $$$a$$$ is available or can be computed efficiently, you can instead do
int ans = lower_bound(a,a+n,1)-a; // get index of leftmost 1
Careful if $$$1$$$ doesn't exist in $$$a$$$.
You can then solve most problems by implementing your own definition of check(). Time complexity is the time complexity of your check() function times $$$\log n$$$. Thanks to marvenlee for inspiring this blog.
Usage examples:
Indeed, there are many other ways of implementing binary search, check out pllk's blog.
