Minimum Operations to Make Array Sum Divisible by K - Leetcode 3512 - Python
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The hook
0:00–0:26Hey everyone, welcome back and let's write some more neat code today. So today, let's solve the problem minimum operations to make array sum divisible by k. So I think this is one of those uh participation credit problems. The idea is that we're given an array of numbers and it's uh relevant to say that every one of those numbers is going to be positive. And basically they say we're allowed to perform this operation any
Structure
Hey everyone, welcome back and let's write some more neat code today. So today, let's solve the problem minimum operations to make array sum divisible by k. So I think this is one of those uh participation credit problems. The idea is that we're given an array of numbers and it's uh relevant to say that every one of those numbers is going to be positive. And basically they say we're allowed to perform this operation any number of times. And so as long as you can interpret this correctly, you can solve the problem pretty quickly. We want to just pick any element and we can decrement it by one. So we want to know what's the minimum number of operations required to make the sum of the entire subarray which remember is going to be positive. The entire sum should be divisible by K. So remember that 0 is also divisible by K. So in the worst case uh like in this example k is equal to 5. It doesn't really matter what the numbers are or how many numbers we have because all we really care about is the sum of the array. So we don't really need to like pick and choose elements and decrement them. It doesn't matter whether we decrement this one or this one. It doesn't matter. So let's say the sum is equal to 14 or something. Well, is that divisible by five? No, it's not.
Watch from 0:00to make it divisible by five, we could either decrement it down to four or down to 10 or we could increment it up to 15. Now, of course, we can't increment it, so we decrement it. So, how do you get like the number of decrements that we would need? Well, you can just take this number, mod it by K, and then you'll get four. What if this number was already less than K? It would have been something like four. Well, you mod it by K, you get four again. And then that means we can decrement this down to zero in four operations. So, what that means is all you do is return the sum of the array and you mod it by K. And I do believe that's the best way to solve the problem. You can see here on the left, it works. It's pretty efficient. Thanks for watching. I'll see you soon.
Watch from 1:19Full transcript
- 0:00
Hey everyone, welcome back and let's write some more neat code today. So today, let's solve the problem minimum operations to make array sum divisible by k. So I think this is one of those uh participation credit problems. The idea is that we're given an array of numbers and it's uh relevant to say that every one of those numbers is going to be
- 0:20
positive. And basically they say we're allowed to perform this operation any number of times. And so as long as you can interpret this correctly, you can solve the problem pretty quickly. We want to just pick any element and we can decrement it by one. So we want to know what's the minimum number of operations required to make the sum of the entire subarray which remember is going to be
- 0:42
positive. The entire sum should be divisible by K. So remember that 0 is also divisible by K. So in the worst case uh like in this example k is equal to 5. It doesn't really matter what the numbers are or how many numbers we have because all we really care about is the
- 1:02
sum of the array. So we don't really need to like pick and choose elements and decrement them. It doesn't matter whether we decrement this one or this one. It doesn't matter. So let's say the sum is equal to 14 or something. Well, is that divisible by five? No, it's not. to make it divisible by five, we could either decrement it down to four or down
- 1:24
to 10 or we could increment it up to 15. Now, of course, we can't increment it, so we decrement it. So, how do you get like the number of decrements that we would need? Well, you can just take this number, mod it by K, and then you'll get four. What if this number was already less than K? It would have been something like four. Well, you mod it by K, you get four again. And then that
- 1:45
means we can decrement this down to zero in four operations. So, what that means is all you do is return the sum of the array and you mod it by K. And I do believe that's the best way to solve the problem. You can see here on the left, it works. It's pretty efficient. Thanks for watching. I'll see you soon.
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