python implementation

def jumpingOnClouds(c):
    # Write your code here
    jumps = 0
    i=0
    c_len=len(c)
    while i < c_len - 1:
        j=2
        while j >0:
            if i+j<=c_len-1:
                if c[i+j]==0:
                    i+=j
                    jumps+=1
                    break
                else:
                    j-=1
            else:
                j-=1
    return jumps

public static int jumpingOnClouds(List<Integer> c) {
// Write your code here

int jumps = 0; int i = 0;

while (i < c.size() - 1) {
    // Prefer jumping 2 steps if possible, otherwise jump 1 step
    if (i + 2 < c.size() && c.get(i + 2) == 0) {
        i += 2;
    } else {
        i += 1;
    }
    jumps++; // Count the jump
}

return jumps;
}

Python 3

def jumpingOnClouds(c):

jump = 0
i = 0
while True:
    if (i+2)<= (len(c)-1) and c[i+2] == 0 :
        jump = jump + 1
        i = i + 2

    elif (i+1)<= len(c)-1:
        i = i + 1
        jump = jump + 1

    else:
        break

return jump

Here is my c++ solution, you can watch the explanation here : https://youtu.be/GNof9B-9CN0

int jumpingOnClouds(vector<int> c) {
    int result = 0, index = 0;
    while(index < c.size() - 1){
        if(c[index + 2] == 1) index++;
        else index+=2;
        result++;
    }
    return result;
}

Python

This solutions separates constants and variables, allowing for faster execution. The max_index and size are constants, allowing us to only measure the length of the array once. For the solution to be optimal, we need to try to jump the furthest we can. Since the question emphasizes that there's always a possible solution, this means that in case there's a dangerous cloud in position + 2 we can always jump to position + 1 and this will be the minimum number of jumps.

def jumpingOnClouds(clouds: list[int])-> int:
    position: int = 0
    jumps: int = 0
    size: int = len(clouds)
    max_index: int = size -1
    
    while position < max_index:
        if(
            position + 2 > max_index or
            clouds[position + 2] == 1
        ):
            position += 1
        else:
            position += 2
        jumps += 1
    
    return jumps