1 00:00:00,560 --> 00:00:01,580 Welcome back. 2 00:00:01,580 --> 00:00:05,720 Let's now code the brute force solution which we discussed in the previous video. 3 00:00:05,720 --> 00:00:08,000 Now for this we are going to write a function. 4 00:00:08,000 --> 00:00:10,850 Let's call it max area brute force. 5 00:00:12,990 --> 00:00:16,110 And this function is going to take in the array. 6 00:00:16,260 --> 00:00:16,560 All right. 7 00:00:16,560 --> 00:00:17,820 So we have array over here. 8 00:00:17,850 --> 00:00:22,920 Now once inside this function we will initialize the area to be equal to zero. 9 00:00:22,920 --> 00:00:23,340 All right. 10 00:00:23,340 --> 00:00:25,770 And then finally we will just return this area. 11 00:00:25,770 --> 00:00:30,180 We will see whether we are getting an area which is greater than what we have over here. 12 00:00:30,180 --> 00:00:33,750 And accordingly we will replace the area with the greater value. 13 00:00:33,750 --> 00:00:34,290 All right. 14 00:00:34,320 --> 00:00:43,020 Now let's say that the area, the array which is given to us is let's say three, seven, five, six, 15 00:00:43,020 --> 00:00:43,650 eight. 16 00:00:43,650 --> 00:00:47,520 And for the same one which we discussed in the video previously. 17 00:00:47,640 --> 00:00:48,060 All right. 18 00:00:48,090 --> 00:00:49,530 Now what are we going to do. 19 00:00:49,530 --> 00:00:55,560 As we discussed we are going to have a variable I and it will go from zero till this eight over here. 20 00:00:55,560 --> 00:00:55,920 Right. 21 00:00:55,920 --> 00:00:57,180 So let's write that over here. 22 00:00:57,180 --> 00:00:57,990 So for. 23 00:00:59,050 --> 00:01:02,650 Let I is equal to zero I less than array. 24 00:01:05,110 --> 00:01:06,370 Dot length. 25 00:01:07,770 --> 00:01:10,080 Minus one I plus plus. 26 00:01:10,770 --> 00:01:16,140 So we have arrived at length minus one because we just want to go up to this element over here. 27 00:01:16,140 --> 00:01:16,530 Right. 28 00:01:16,560 --> 00:01:20,550 Now once we are inside this for loop we are going to have another for loop. 29 00:01:20,550 --> 00:01:25,020 So let's say for let j is equal to I plus one right. 30 00:01:25,020 --> 00:01:28,290 So we're going to start from the next element after I. 31 00:01:28,320 --> 00:01:28,740 Right. 32 00:01:28,740 --> 00:01:32,730 And then we are going to check up to j less than array dot length. 33 00:01:34,880 --> 00:01:36,890 And then we are going to say J plus, plus. 34 00:01:38,900 --> 00:01:39,380 All right. 35 00:01:39,380 --> 00:01:40,760 Now, what are we going to do? 36 00:01:40,790 --> 00:01:44,420 We're going to find the minimum height at these two indices. 37 00:01:44,420 --> 00:01:44,780 Right. 38 00:01:44,780 --> 00:01:46,940 So let's say the height. 39 00:01:46,940 --> 00:01:48,980 Let's say let height. 40 00:01:49,930 --> 00:01:51,130 Or we can say const height. 41 00:01:51,130 --> 00:01:56,230 At this point, const height is equal to math dot min. 42 00:01:56,350 --> 00:02:01,930 And then we're going to find the minimum right height at these two indices at I and j. 43 00:02:01,930 --> 00:02:04,300 So let's say min array at I. 44 00:02:06,510 --> 00:02:07,470 And RJ. 45 00:02:09,400 --> 00:02:10,690 So that will give us the height. 46 00:02:10,690 --> 00:02:11,140 Right. 47 00:02:11,140 --> 00:02:12,520 And what about the width. 48 00:02:12,520 --> 00:02:13,930 So const width. 49 00:02:14,950 --> 00:02:16,480 Will be equal to. 50 00:02:17,280 --> 00:02:18,720 J minus I. 51 00:02:18,750 --> 00:02:19,020 Right. 52 00:02:19,020 --> 00:02:26,640 So if j is equal to one and I is equal to zero, then we know that the width is zero one -0 which is 53 00:02:26,640 --> 00:02:27,180 equal to one. 54 00:02:27,180 --> 00:02:29,670 So that's j minus I over here. 55 00:02:30,150 --> 00:02:32,430 So we have got the height and width right. 56 00:02:32,460 --> 00:02:36,180 Now we can just find the area by multiplying the height and width. 57 00:02:36,180 --> 00:02:42,870 And then we are going to say area is equal to math dot max because we want the maximum area right among 58 00:02:42,870 --> 00:02:46,590 the current area and height dot height into width. 59 00:02:49,350 --> 00:02:49,770 All right. 60 00:02:49,770 --> 00:02:53,250 So this will give us the maximum area. 61 00:02:53,250 --> 00:02:53,400 Right. 62 00:02:53,400 --> 00:02:56,940 So the maximum area will always be stored into area. 63 00:02:56,940 --> 00:02:58,560 And then we can just return area. 64 00:02:58,560 --> 00:02:59,310 So that's it right. 65 00:02:59,310 --> 00:03:00,600 So we're done with our function. 66 00:03:00,600 --> 00:03:02,580 Now let's go ahead and test this. 67 00:03:02,580 --> 00:03:04,500 Now to test it's easy. 68 00:03:04,500 --> 00:03:09,180 Let's just pass this array over here because we have already discussed this in the previous video. 69 00:03:09,180 --> 00:03:13,230 So the expected maximum area is 21 right. 70 00:03:13,230 --> 00:03:16,650 So when we take seven and eight so that's seven into three. 71 00:03:16,650 --> 00:03:20,790 So let's just call the function so max area brute force. 72 00:03:20,790 --> 00:03:23,820 And we're going to pass in this array over here. 73 00:03:24,760 --> 00:03:26,860 Now let's see what we get as the output. 74 00:03:26,860 --> 00:03:28,420 So let's just run this. 75 00:03:28,420 --> 00:03:31,690 And the output that we're getting is 21 which is correct. 76 00:03:31,690 --> 00:03:32,020 Right. 77 00:03:32,020 --> 00:03:33,910 So yes our function is working. 78 00:03:33,910 --> 00:03:35,770 Now let's take another example. 79 00:03:35,770 --> 00:03:38,740 Also let's say we are passing an array where it's obvious. 80 00:03:38,740 --> 00:03:41,890 Let's say we have the largest values at the two ends. 81 00:03:41,890 --> 00:03:43,930 So 9123. 82 00:03:43,930 --> 00:03:45,610 And then let's say we have ten right. 83 00:03:45,610 --> 00:03:48,370 So what is the expected area. 84 00:03:48,370 --> 00:03:51,310 In this case it should be the height is nine right. 85 00:03:51,310 --> 00:03:53,080 It should be the minimum of nine and ten. 86 00:03:53,080 --> 00:03:55,840 And the width will be 123 and four. 87 00:03:55,840 --> 00:03:57,400 So we should get 36 right. 88 00:03:57,400 --> 00:03:59,920 So yes we are getting 36 as the output over here. 89 00:03:59,920 --> 00:04:02,020 So yes this function is working. 90 00:04:02,020 --> 00:04:08,470 Now let's take a sample input and walk through the code and relook at the time and space complexity. 91 00:04:08,470 --> 00:04:10,690 Remember in the interview this is an important step. 92 00:04:10,690 --> 00:04:15,040 Once you have written the code, you always have to take some sample input and walk through the code. 93 00:04:15,040 --> 00:04:16,870 So let's go ahead and look at the code. 94 00:04:16,870 --> 00:04:17,680 Walk through now.