1 00:00:00,240 --> 00:00:01,350 Welcome back, everyone. 2 00:00:01,710 --> 00:00:06,840 In the last video, I mentioned that there is a further optimization we can make when we realized that 3 00:00:06,840 --> 00:00:13,110 we're only leveraging the last two stored values inside of our DP as we iterate through AI to end. 4 00:00:13,590 --> 00:00:19,350 What we can do now is just replace this deep scaling array with two variables instead. 5 00:00:19,860 --> 00:00:22,860 So to start, I'm just going to erase all of the code. 6 00:00:22,860 --> 00:00:27,390 That is irrelevant because this is the part of it that we're going to rewrite. 7 00:00:27,720 --> 00:00:30,980 I'm also going to get rid of the space and time complexity here. 8 00:00:30,990 --> 00:00:32,580 Just remember that it was O of end. 9 00:00:33,420 --> 00:00:39,750 Next, what we're going to do is just think about the two variables we need to initialize when we start 10 00:00:39,750 --> 00:00:40,590 this value. 11 00:00:41,100 --> 00:00:45,120 DP one, which is what I'm going to call the first variable. 12 00:00:46,080 --> 00:00:53,070 Is going to actually be equal to cost at zero because this is the starting variable that we need to 13 00:00:53,070 --> 00:00:53,470 store. 14 00:00:54,240 --> 00:01:00,320 Secondly, then you can assume that DP two is going to be equal to cost at one. 15 00:01:01,110 --> 00:01:05,760 Now that we've initialized our two DP variables, we can just run through the iteration like we did 16 00:01:05,760 --> 00:01:07,080 before here. 17 00:01:07,080 --> 00:01:12,900 We're going to use the same for loop, except we're going to initialize at two instead of zero, because 18 00:01:12,900 --> 00:01:20,040 essentially what we did when we set up one in two was we cast the variables as the first two instances 19 00:01:20,040 --> 00:01:22,320 of I we don't need to generate DP's for them. 20 00:01:22,330 --> 00:01:24,970 That's what these two lines have literally done for us. 21 00:01:25,350 --> 00:01:29,070 So with AI starting at two, we're going to go up to end the same way. 22 00:01:29,490 --> 00:01:36,180 And I plus plus and now inside, what we want to do is generate the current value for AI. 23 00:01:36,750 --> 00:01:40,860 So here it's going to utilize the two values at our DP. 24 00:01:41,100 --> 00:01:45,510 So it's going to be cost at AI plus math dot min. 25 00:01:48,800 --> 00:01:49,700 DP won. 26 00:01:50,570 --> 00:01:51,590 And DP to. 27 00:01:53,640 --> 00:02:00,660 Once we've generated and figured out what this current value is for, I, we need to replace our deep 28 00:02:00,690 --> 00:02:01,120 one. 29 00:02:01,800 --> 00:02:08,760 So first, what we're going to do is say that DPE one is equal to DPE two, because as we know from 30 00:02:08,760 --> 00:02:16,350 the tree, when we use DPE one and two to generate the current value, we're going to use these two 31 00:02:16,350 --> 00:02:21,150 now to generate this value, which means that this DPE one value is no longer needed. 32 00:02:21,380 --> 00:02:24,510 Instead, this value is going to be the new DPE one. 33 00:02:24,690 --> 00:02:26,130 This value we just generate. 34 00:02:26,130 --> 00:02:30,230 It is the new DPE two or just keeping track of the two most recent values. 35 00:02:30,600 --> 00:02:38,460 So that's why DPE one is now equal to TPE two and two is equal to our current value. 36 00:02:39,510 --> 00:02:45,810 And that's all we need to do, just like that, we've generated and built up the same deep and of course, 37 00:02:45,810 --> 00:02:51,200 the final answer that we return is just going to be the men of these two values. 38 00:02:51,390 --> 00:02:56,700 So we're just going to return math to men between DPE one and deep two. 39 00:02:58,420 --> 00:03:02,770 Once again, the reason why this happens is because the last two values that we've sought should be 40 00:03:02,770 --> 00:03:07,300 the deep values for these last two variables inside of our staircase. 41 00:03:08,340 --> 00:03:09,940 And this is the optimization. 42 00:03:10,530 --> 00:03:17,130 Now, let's compare the space and time, if we look at our time complexity, first we notice that there's 43 00:03:17,130 --> 00:03:18,860 no real optimization here. 44 00:03:18,870 --> 00:03:23,940 We're still going to loop over the majority of N, but we're only going to do it once still. 45 00:03:24,270 --> 00:03:31,560 So here it's still O of N, but when we look at our space complexity, we have no scaling data structures 46 00:03:31,560 --> 00:03:32,010 anymore. 47 00:03:32,340 --> 00:03:39,120 Instead, we just initialize to static variables, which means that we ended up with an O of one space 48 00:03:39,120 --> 00:03:45,480 complexity, which is an amazing improvement over are even very first recursive solution. 49 00:03:45,630 --> 00:03:49,590 If you think back, they were both two to the end here. 50 00:03:49,590 --> 00:03:56,850 We've brought this down to of non-linear and of one that is incredible performance, constant space 51 00:03:56,850 --> 00:03:58,050 and linear time. 52 00:03:58,170 --> 00:04:00,210 That is one of the best possible solutions. 53 00:04:00,420 --> 00:04:06,600 And this is why dynamic programming is so powerful as an algorithmic paradigm. 54 00:04:06,900 --> 00:04:12,240 If you think about it, the very nature of the fact that we have to generate all the possible optional 55 00:04:12,240 --> 00:04:17,250 solutions that could exist for a given question and then pick the best one from amongst them, that 56 00:04:17,250 --> 00:04:21,210 by default sounds like a very computationally expensive endeavor. 57 00:04:21,510 --> 00:04:28,380 But through dynamic programming, we're able to bring that down to such a drastically performance solution. 58 00:04:29,260 --> 00:04:34,120 So here is the key that you have to remember, there are so many steps of dynamic programming, the 59 00:04:34,120 --> 00:04:39,220 first one begins with identifying the recurrence relation formula that you can then use to build the 60 00:04:39,220 --> 00:04:45,130 recursive function from the recursive function you want to build at top down and determine if there 61 00:04:45,130 --> 00:04:48,280 is an optimization using memorisation that can help you. 62 00:04:48,580 --> 00:04:55,090 Once you see that full top down solution, it's easy for you to extrapolate out the bottom up solution 63 00:04:55,270 --> 00:04:59,620 because you know that if you're going top down, there's definitely a way to go bottom up. 64 00:04:59,620 --> 00:05:04,660 As long as you analyze the correct base cases that you might have, most of the time you'll have these 65 00:05:04,660 --> 00:05:08,200 base cases and they'll help you think about that bottom up approach. 66 00:05:08,620 --> 00:05:13,990 Once you've understood what that is, you figure out the bottom up tree is and then you derive it ideally. 67 00:05:14,320 --> 00:05:22,030 Then you look at your DPI and see if there is any wasted usage of space inside of your scaling DPI. 68 00:05:22,300 --> 00:05:28,210 If there is, you reduce it to the parts that you only need and you're able to bring down to at least 69 00:05:28,210 --> 00:05:32,750 the next level of space complexity with our own notation. 70 00:05:33,580 --> 00:05:36,510 This is dynamic programming and this is the power of it. 71 00:05:37,270 --> 00:05:44,110 If this whole process seems a little bit still kind of vague to you, don't worry, we're going to practice 72 00:05:44,110 --> 00:05:44,850 a lot more. 73 00:05:45,130 --> 00:05:49,720 So just keep this whole step by step, stage by stage process in your mind. 74 00:05:50,260 --> 00:05:57,640 And always remember that with your interviews, you may not require going through the entire dynamic 75 00:05:57,640 --> 00:06:01,470 programming step to get to this final optimized solution. 76 00:06:01,750 --> 00:06:07,270 A lot of the times you can just finish the full top down solution with the memorization. 77 00:06:07,810 --> 00:06:12,840 The bottom up is kind of hard to get to within the tight time frame of an interview question. 78 00:06:13,120 --> 00:06:18,940 However, if you're able to articulate what that next step of going from top down to bottom up brings 79 00:06:18,940 --> 00:06:25,090 in terms of its benefits and even just understanding the process of going from that top down tree to 80 00:06:25,090 --> 00:06:30,310 this bottom up tree, if you're able to just articulate this stage of it, where you logically thinking 81 00:06:30,310 --> 00:06:36,640 about it but not coding it out, that is oftentimes enough for your interviewer to know that you could 82 00:06:36,640 --> 00:06:39,250 have solved the problem had you had more time. 83 00:06:39,400 --> 00:06:43,900 And that is often enough for them to pass you through the interview so deeply. 84 00:06:43,900 --> 00:06:49,300 Understand the process that we took to break down this dynamic programming problem and we'll get more 85 00:06:49,300 --> 00:06:51,250 practice with the next problem.