WEBVTT 00:01.300 --> 00:01.810 Hi everyone. 00:01.810 --> 00:03.070 Welcome back to this tutorial. 00:03.070 --> 00:10.780 In this tutorial, we'll code and learn how to perform the min operation of a list of integers in a 00:10.780 --> 00:11.440 stream. 00:11.440 --> 00:15.370 So here we talked about the max in the previous tutorial. 00:15.370 --> 00:17.410 I'm going to use the same example here. 00:17.440 --> 00:22.840 The first thing is I want to show the difference between with the identity and without the identity 00:22.840 --> 00:23.560 value. 00:24.220 --> 00:28.450 I'm going to copy the same code and then paste it here. 00:28.480 --> 00:31.030 It's going to be named as min min value. 00:31.450 --> 00:33.640 I'm going to comment this whole max part. 00:33.670 --> 00:34.750 I don't want this. 00:35.590 --> 00:38.410 Because I don't I'm not going to use this as part of this tutorial. 00:38.560 --> 00:41.320 Now, let me put this in presentation mode. 00:46.210 --> 00:50.110 Then here we are going to talk about the min value so far, the max value. 00:50.140 --> 00:54.910 What we did was we used the greater than for this. 00:54.940 --> 00:58.210 We are just going to change this one to less than. 00:58.570 --> 00:59.380 Let's talk about it. 00:59.380 --> 01:00.850 What happens behind the scenes. 01:00.850 --> 01:08.080 So here six is going to be assigned against the Y variable here and X is going to hold the result of 01:08.080 --> 01:08.680 it, right? 01:08.680 --> 01:13.810 So for the first time, what it is going to do, it is going to assign zero. 01:14.020 --> 01:21.400 That is the initial value against the X variable and then Y is going to hold six for the first element 01:21.400 --> 01:26.950 in the string because we know string is going to pass the elements one by one and then is zero less 01:26.950 --> 01:27.850 than six. 01:28.270 --> 01:32.110 So here is a condition which we which we have is a zero less than six. 01:32.110 --> 01:32.440 Yes. 01:32.470 --> 01:37.360 In that case X will be assigned with zero value for the next element. 01:37.540 --> 01:44.380 Seven is again, seven is set against the y variable is zero less than seven. 01:44.380 --> 01:45.070 Yes. 01:45.070 --> 01:49.100 And then it is going to assign the zero as a minimum value. 01:49.100 --> 01:54.560 It is going to continue until all the elements in the stream are passed as part of the iteration. 01:54.560 --> 01:59.300 What I'm going to do, I'm going to call this method and then check what is the result that it is going 01:59.300 --> 01:59.810 to give. 02:00.470 --> 02:05.210 So int min value equal to. 02:06.420 --> 02:08.850 Find min value and then pass integer list. 02:10.790 --> 02:14.870 You're going to be surprised by looking at the result because. 02:16.000 --> 02:17.710 We'll get to know about that in a bit. 02:20.800 --> 02:21.610 Run this. 02:23.040 --> 02:28.800 So the output that we're expecting is six, but the result that we got is zero. 02:28.830 --> 02:29.650 Why? 02:29.670 --> 02:34.800 If we take a look at it, we have the default value, which is set to zero. 02:34.890 --> 02:39.750 That is going to be the least value compared to all the elements in the string. 02:39.870 --> 02:40.530 Right. 02:40.530 --> 02:45.660 So this is something which we have to keep in our mind when you are calculating the min value. 02:45.690 --> 02:48.960 You should never, ever use the default value. 02:49.160 --> 02:53.220 So I'm going to remove this and then let's change this one to optional. 02:55.830 --> 02:56.730 Same concept. 02:57.340 --> 03:00.540 And then I'm going to copy this and put it here. 03:01.970 --> 03:03.650 I'll be minvalue. 03:04.910 --> 03:05.450 Optional. 03:06.430 --> 03:08.560 And then now the drill. 03:09.840 --> 03:12.150 Going to print the mean value optional. 03:15.370 --> 03:23.380 Here the f min value dot is present, then I will get the value and then printed this out. 03:26.660 --> 03:29.360 The minimum value is. 03:33.220 --> 03:35.740 Minvalue optional dot get. 03:36.640 --> 03:40.990 Good action is else going to print? 03:43.700 --> 03:44.720 No input is. 03:48.220 --> 03:50.330 Now let's run this and check. 03:50.350 --> 03:53.530 So we are we're calling the find min value. 03:53.530 --> 04:00.760 It is taking a list is iterating, and then it is going to apply each and every value in the stream 04:00.760 --> 04:05.320 to the Y variable, and X is going to hold the result as the. 04:07.270 --> 04:08.140 X variable. 04:08.230 --> 04:11.180 So in here, let's run this and check. 04:11.200 --> 04:14.080 So the value that I'm expecting is six. 04:16.560 --> 04:17.580 We take a look at it. 04:17.610 --> 04:18.620 The value is six. 04:18.630 --> 04:20.400 Let's say I'm passing an empty list. 04:20.430 --> 04:21.510 What will happen? 04:22.350 --> 04:24.150 Empty list is going to. 04:25.630 --> 04:28.090 Print the result as no input is passed. 04:28.540 --> 04:29.030 Take a look at it. 04:29.050 --> 04:31.720 No input is passed and the optional value is. 04:33.430 --> 04:41.290 So this is one of the approach of calculating the min value using the reduce function. 04:42.190 --> 04:48.520 So this is all about how to calculate the max and min value from a stream of integers. 04:48.700 --> 04:53.590 You can just do it for any kind of other number types. 04:53.860 --> 04:55.150 Double long. 04:55.300 --> 04:56.500 The same approach. 04:57.600 --> 04:59.610 But thus we came to the end of this tutorial. 04:59.640 --> 05:00.720 Thank you for watching.