1 00:00:01,500 --> 00:00:02,110 Hello everyone. 2 00:00:02,150 --> 00:00:03,140 Welcome to the session. 3 00:00:03,140 --> 00:00:06,110 So today we will solve one more problem with the help of the occasion. 4 00:00:06,140 --> 00:00:08,840 So the name of the problem is currency it. 5 00:00:09,290 --> 00:00:11,600 OK so what is the problem. 6 00:00:11,600 --> 00:00:17,760 So the problem is very simple given a number let's say the number is 1 0 3 2 0. 7 00:00:17,840 --> 00:00:20,300 So we have to calculate how many zeros at present. 8 00:00:20,330 --> 00:00:22,420 So in this case our output will be two. 9 00:00:22,440 --> 00:00:22,720 Okay. 10 00:00:22,730 --> 00:00:26,490 Because there are two zeros okay. 11 00:00:26,500 --> 00:00:30,340 Now let's take one more example of the value of and this let's say 123. 12 00:00:30,340 --> 00:00:33,390 Then output will be zero because there are no 0 present. 13 00:00:33,760 --> 00:00:34,210 OK. 14 00:00:34,270 --> 00:00:39,420 So I hope you've got the problem we just have to count the number of zeros present inside the number. 15 00:00:39,860 --> 00:00:42,390 OK now how we can solve this problem. 16 00:00:42,400 --> 00:00:45,700 So basically what we have to do we will go to each and every digit. 17 00:00:46,180 --> 00:00:47,940 We will go to each and every digit. 18 00:00:47,950 --> 00:00:52,420 And if the digit is 0 we will do our answer plus plus. 19 00:00:52,420 --> 00:00:52,650 Okay. 20 00:00:52,660 --> 00:00:56,200 So this was the approach of when we are using for loop of divine loop. 21 00:00:56,200 --> 00:00:56,730 OK. 22 00:00:56,890 --> 00:01:00,190 Now we will use a similar approach if we are using recursion also. 23 00:01:00,190 --> 00:01:01,400 So what we will do. 24 00:01:01,450 --> 00:01:05,850 So given that the Model T the number is 1 0 2 2 0 only. 25 00:01:05,920 --> 00:01:07,060 So what I will do. 26 00:01:07,060 --> 00:01:10,180 So I will tell that equation. 27 00:01:10,180 --> 00:01:12,820 Give me how many zeros at present in this Malone. 28 00:01:12,910 --> 00:01:16,370 So the equation will give me that answer for this is one. 29 00:01:16,420 --> 00:01:16,630 Okay. 30 00:01:16,660 --> 00:01:21,550 Because there are 1 0 present then what we will do we will check the last digit. 31 00:01:21,820 --> 00:01:22,380 OK. 32 00:01:22,450 --> 00:01:27,730 We will check this last digit OK. 33 00:01:27,740 --> 00:01:29,060 So this last digit is it. 34 00:01:29,070 --> 00:01:29,830 So what do we do. 35 00:01:29,840 --> 00:01:31,760 We will do one plus one. 36 00:01:31,820 --> 00:01:32,350 OK. 37 00:01:32,510 --> 00:01:34,790 Since the last budget is it what we will do plus one. 38 00:01:34,790 --> 00:01:37,730 So the answer will become to OK. 39 00:01:37,760 --> 00:01:40,670 So this is how this code will work if we want to. 40 00:01:40,670 --> 00:01:43,120 So this will be now. 41 00:01:43,240 --> 00:01:45,410 This is the function count zeros. 42 00:01:45,730 --> 00:01:48,010 It will take a number and as argument. 43 00:01:48,010 --> 00:01:49,180 What I will do. 44 00:01:49,180 --> 00:01:55,490 So this cone zeros function concludes the count convert the number of zeros present inside the number. 45 00:01:55,480 --> 00:01:57,690 And so I will give the smaller input. 46 00:01:58,000 --> 00:02:00,610 So it will give me how many zeros at present in the number. 47 00:02:00,610 --> 00:02:01,580 And by 10. 48 00:02:01,630 --> 00:02:05,520 So this is basically and by 10 OK. 49 00:02:05,530 --> 00:02:07,660 So this will be our problem. 50 00:02:07,690 --> 00:02:09,880 This will be done stuff with a similar problem. 51 00:02:09,880 --> 00:02:12,020 And then we will check the last digit. 52 00:02:12,370 --> 00:02:17,200 OK so we will check if the last digit in this case it is 0 here. 53 00:02:17,230 --> 00:02:23,380 So if the last digit is 0 what we will do we will add plus 1 to our answer. 54 00:02:23,410 --> 00:02:26,770 So this was our smaller answer. 55 00:02:26,770 --> 00:02:27,540 So what do we do. 56 00:02:27,550 --> 00:02:34,330 We will lose more Laron said plus plus or we can write a small lines plus one in the each part. 57 00:02:34,330 --> 00:02:38,780 You will simply there done smaller answer or gave the last. 58 00:02:38,780 --> 00:02:39,630 It is not as you don't. 59 00:02:39,650 --> 00:02:40,880 We will not do plus one. 60 00:02:41,150 --> 00:02:43,930 So we'll simply return small answer. 61 00:02:45,050 --> 00:02:45,380 OK. 62 00:02:45,920 --> 00:02:47,350 And the latitude is it. 63 00:02:47,360 --> 00:02:49,380 We will dance Malaysia plus 1. 64 00:02:49,400 --> 00:02:51,340 Let's see how this code will work. 65 00:02:51,410 --> 00:02:52,060 So. 66 00:02:52,340 --> 00:02:53,980 OK so 1 0 2 2 0. 67 00:02:53,990 --> 00:02:55,220 Take this example only. 68 00:02:55,820 --> 00:03:00,390 So I have this example 1 0 3 2 1 0. 69 00:03:00,560 --> 00:03:03,210 So make the Parliament a smaller part. 70 00:03:03,250 --> 00:03:04,820 With qualification on this part. 71 00:03:05,400 --> 00:03:05,980 OK. 72 00:03:06,060 --> 00:03:10,340 So this will become 1 0 3 2. 73 00:03:10,440 --> 00:03:14,360 Then again call that equation a smaller part. 74 00:03:14,390 --> 00:03:16,710 So this will become 1 0 3. 75 00:03:17,090 --> 00:03:18,980 Again call that equation a smaller part. 76 00:03:19,010 --> 00:03:23,150 So this will become 10 again called that equation on the smaller part. 77 00:03:23,750 --> 00:03:27,650 So this will become one again called education the smaller part. 78 00:03:27,920 --> 00:03:29,470 So this part is 0. 79 00:03:29,510 --> 00:03:31,290 So I'm calling 1 0. 80 00:03:31,340 --> 00:03:35,360 Now this will become our base case. 81 00:03:35,370 --> 00:03:36,180 So what will happen. 82 00:03:36,180 --> 00:03:38,900 I will return 0 okay. 83 00:03:38,920 --> 00:03:41,510 Now we will check the last digit losses of this one. 84 00:03:41,560 --> 00:03:44,260 So simpler than the smaller answer. 85 00:03:44,270 --> 00:03:45,480 We will check the last digit. 86 00:03:45,490 --> 00:03:46,710 Another last digit is zero. 87 00:03:46,720 --> 00:03:50,170 So what do we do smaller on the set plus 1. 88 00:03:50,170 --> 00:03:52,720 So we will return 0 plus 1 which is 1. 89 00:03:52,720 --> 00:03:53,910 Now check the last digit. 90 00:03:53,920 --> 00:03:55,550 It is not zero. 91 00:03:55,600 --> 00:03:59,110 So it is simpler than the smaller answer which is one only. 92 00:03:59,110 --> 00:04:01,470 Okay now check the last budget which is 2. 93 00:04:01,740 --> 00:04:05,190 So we will simply write down the smaller answer. 94 00:04:05,200 --> 00:04:06,400 Now check the last digit. 95 00:04:06,430 --> 00:04:07,330 So this is zero. 96 00:04:07,480 --> 00:04:08,630 So we will add plus 1. 97 00:04:08,980 --> 00:04:12,660 So this will become one plus one which is 2 so too is our answer. 98 00:04:12,710 --> 00:04:13,220 Okay. 99 00:04:13,270 --> 00:04:15,130 So that's all very simple good. 100 00:04:15,130 --> 00:04:15,750 Okay. 101 00:04:15,910 --> 00:04:18,260 Now let us write the code. 102 00:04:18,550 --> 00:04:20,660 So what was the return type. 103 00:04:20,710 --> 00:04:23,180 We all know that you don't type will be integer. 104 00:04:23,440 --> 00:04:28,160 Let's say the name of the function is count zeros what it will take. 105 00:04:28,160 --> 00:04:29,130 It will take a number. 106 00:04:29,130 --> 00:04:32,790 And this argument again two simple steps. 107 00:04:33,000 --> 00:04:34,400 First the base case. 108 00:04:34,410 --> 00:04:35,900 Now the base case is very simple. 109 00:04:35,910 --> 00:04:38,580 If the value of an 0. 110 00:04:38,610 --> 00:04:39,630 We will return 0 111 00:04:42,610 --> 00:04:43,530 if the value. 112 00:04:43,540 --> 00:04:50,230 Now it's time for the recursive case so what is a recursive case. 113 00:04:50,230 --> 00:04:53,240 So the recursive case what we talked about. 114 00:04:53,260 --> 00:04:55,750 Okay so what is a recursive case. 115 00:04:55,750 --> 00:04:57,490 So small answer 116 00:05:00,930 --> 00:05:05,520 calculate the zeros present in the number and by then. 117 00:05:05,780 --> 00:05:06,080 Okay. 118 00:05:06,110 --> 00:05:12,320 So we are concluding the Zulus peasant dinner then a mutt and Martin and then our calculation part 119 00:05:15,240 --> 00:05:16,290 so what is that calculation. 120 00:05:16,280 --> 00:05:19,940 But we will use a friend and so if the last digit. 121 00:05:19,960 --> 00:05:22,560 OK so first of all let's calculate the last digit 122 00:05:25,470 --> 00:05:32,940 so the last digit is and more than so the last digit as and more ten. 123 00:05:33,060 --> 00:05:34,420 Now we will check. 124 00:05:34,500 --> 00:05:40,280 So if the last digit equals equals zero what do we do. 125 00:05:40,280 --> 00:05:50,010 So our answer will become we will have done one plus small onset otherwise in the else part we will 126 00:05:50,010 --> 00:05:52,330 simply write down our smaller answer. 127 00:05:52,440 --> 00:05:52,910 Okay. 128 00:05:55,850 --> 00:05:59,850 It will seem better than our small my answer okay. 129 00:06:00,020 --> 00:06:03,120 So this is very simple called. 130 00:06:03,310 --> 00:06:05,050 Now let's call this function 131 00:06:11,410 --> 00:06:14,710 so let's say the value is 1 0 3 2 0 only. 132 00:06:14,920 --> 00:06:16,840 Okay. 133 00:06:17,140 --> 00:06:21,130 Okay so let's make Okay 1 0 2 2 0 only. 134 00:06:21,130 --> 00:06:22,380 Okay. 135 00:06:22,750 --> 00:06:30,150 Now let's turn our file sought out put this coming out to be true because there are two zeros present 136 00:06:30,370 --> 00:06:30,850 and then. 137 00:06:30,880 --> 00:06:34,230 But okay now let's. 138 00:06:34,290 --> 00:06:38,700 Now we were bright in our code and we will make the diagram. 139 00:06:38,700 --> 00:06:43,260 Okay so one more thing in this question the value of and will be good denoted close to one. 140 00:06:43,260 --> 00:06:45,240 Okay so this is given. 141 00:06:45,270 --> 00:06:45,600 Okay. 142 00:06:45,630 --> 00:06:50,620 This is then put this is the constraint the value of N will be given will be good in order to win. 143 00:06:50,820 --> 00:06:51,230 Okay. 144 00:06:51,390 --> 00:06:52,640 Now let's say I didn't a code. 145 00:06:53,490 --> 00:06:55,560 So let's take a different example. 146 00:06:56,060 --> 00:07:03,130 Okay let's say the value is 2 0 0 2 Let's say the number is closer as it would do. 147 00:07:03,130 --> 00:07:04,740 So our answer should be 2. 148 00:07:04,780 --> 00:07:05,270 Okay. 149 00:07:06,400 --> 00:07:07,270 Now what will happen. 150 00:07:07,300 --> 00:07:09,610 So this is not close to zero. 151 00:07:09,610 --> 00:07:10,470 So we will. 152 00:07:10,720 --> 00:07:13,260 We will call this function the smaller part. 153 00:07:13,300 --> 00:07:16,630 So I am calling this function on 200. 154 00:07:16,660 --> 00:07:19,310 This function will wait at line number 10. 155 00:07:19,450 --> 00:07:21,490 Again this function will be returned line number 10. 156 00:07:21,520 --> 00:07:26,220 We will call in 20 dysfunctional wait at line number 10. 157 00:07:26,230 --> 00:07:30,150 We will call on to this function will wait at line number 10. 158 00:07:30,310 --> 00:07:32,170 It will call on 0. 159 00:07:32,170 --> 00:07:35,090 Okay so we will hit the base case here okay. 160 00:07:35,110 --> 00:07:39,740 We are getting the base case it is attending 0 it is bending 0. 161 00:07:39,750 --> 00:07:42,390 So our small answer is 0. 162 00:07:42,660 --> 00:07:44,100 And then the last digit. 163 00:07:44,190 --> 00:07:44,930 And Martin. 164 00:07:44,940 --> 00:07:48,120 So the last digit is two more 10 which is 2. 165 00:07:48,120 --> 00:07:50,180 Okay so 2 is not too close to zero. 166 00:07:50,190 --> 00:07:52,140 So a little simpler than the small answer. 167 00:07:52,140 --> 00:07:53,680 So we will simply order 10 0. 168 00:07:53,940 --> 00:07:54,790 Okay. 169 00:07:54,780 --> 00:07:58,880 West Ambler Dunning zero cell is my line set is zero. 170 00:07:58,950 --> 00:08:01,860 Then again we will conclude the last digit. 171 00:08:01,920 --> 00:08:02,260 Okay. 172 00:08:02,280 --> 00:08:04,480 This function always waiting outside Number 10. 173 00:08:04,500 --> 00:08:06,140 Now it will come to light number 13. 174 00:08:06,150 --> 00:08:08,800 We will conclude the last digit and Martin. 175 00:08:08,860 --> 00:08:10,490 So a lot of it is zero. 176 00:08:10,650 --> 00:08:14,640 If the last digit is we don't we are returning 1 plus small answer. 177 00:08:14,640 --> 00:08:17,140 So 0 plus 1. 178 00:08:17,420 --> 00:08:20,550 Sylvia Dunning when now the spotlight sort of becomes human. 179 00:08:20,850 --> 00:08:24,870 Okay again we will check the last digit losses to zero. 180 00:08:25,380 --> 00:08:27,020 So we will return when bless. 181 00:08:27,060 --> 00:08:29,020 Small answer which is 2. 182 00:08:29,060 --> 00:08:29,290 Okay. 183 00:08:29,310 --> 00:08:31,450 So our small answer is 2. 184 00:08:31,710 --> 00:08:34,650 Now the last digit is 2 2 is not close to zero. 185 00:08:34,680 --> 00:08:36,650 So will simpler than the small answer. 186 00:08:36,680 --> 00:08:38,790 So will simply return to. 187 00:08:38,910 --> 00:08:39,180 Okay. 188 00:08:39,180 --> 00:08:40,590 And this is our answer. 189 00:08:40,800 --> 00:08:41,300 Okay. 190 00:08:41,400 --> 00:08:43,590 So I will call it is working fine. 191 00:08:43,590 --> 00:08:44,100 Okay. 192 00:08:44,160 --> 00:08:46,370 So I hope you understand this problem also. 193 00:08:46,770 --> 00:08:47,150 Okay. 194 00:08:47,190 --> 00:08:48,690 If you have any doubt you can ask me 195 00:08:55,390 --> 00:08:57,220 okay so I hope you understand this problem. 196 00:08:57,220 --> 00:08:59,510 So if you have any doubt you can definitely ask me. 197 00:08:59,580 --> 00:08:59,990 Okay. 198 00:09:00,160 --> 00:09:00,540 Thank you.