java 2 liner

the delta for kelly has to be > 0, or she will never surpass

the difference / delta are the required days

adding 1 to difference to account for a day 0 case

public static int minNum(int samDaily, int kellyDaily, int difference) {
    int kellyDelta = kellyDaily - samDaily;
    return kellyDelta <= 0 ? -1 : (int) Math.ceil((double) (difference + 1) / kellyDelta);
}

**Binary Search approach ** bool canSurpass(int samDaily, int kellyDaily, int difference, int days) { int samSolved = difference + days * samDaily; int kellySolved = days * kellyDaily; return kellySolved > samSolved; }

int minNum(int samDaily, int kellyDaily, int difference) { if (samDaily >= kellyDaily) { return -1; }

int low = 0;
int high = difference / (kellyDaily - samDaily) + 1;

while (low <= high) {
    int mid = low + (high - low) / 2;
    if (canSurpass(samDaily, kellyDaily, difference, mid)) {
        high = mid - 1;
    } else {
        low = mid + 1;
    }
}

if (canSurpass(samDaily, kellyDaily, difference, low)) {
    return low;
} else {
    return -1;
}

}****

JAVA Solution:

public static int minNum(int samDaily, int kellyDaily, int difference) {
    // Write your code here
    
    int samsolved = difference + samDaily;
    int kellysolved = kellyDaily;
    int count=1;
    
    if(kellyDaily==samDaily) return -1;
      
    while(samsolved-kellysolved>=0){
        samsolved=samsolved+samDaily;
        kellysolved=kellysolved+kellyDaily;
        count++;
    }
    return count;
    }

How the problem is constructed at first is a bit confusing, but once you get the general idea. That's the time you realize it's not so difficult.

You only need to understand and get only the minimum number of days that personA will exceed number of problems solve against personB.