Binary Search problem | Part 46 | DSA in java in telugu | Engineering Animuthyam

[@Y.NAGABHUSHANAYEDDULA]

class Solution { public int maximumCount(int[] nums) { int negcount=0; int poscount=0; int ans=0; for(int i=0;i

[@unknownuser-h9r]

macha we can also consider leftmost of zero and right most of zero. i sovled in this way.

[@ajayramavath8413]

Did this in my first try ->Found left most of zero and right most of zero and solved : public static int maximumCount(int[] nums) { int l = left(nums); int r = right(nums); return Math.max(l, nums.length - r - 1); } private static int left(int[] nums) { int start = 0; int end = nums.length - 1; int target = 0; while (start = target) { end = mid - 1; } else { start = mid + 1; } } return start; } private static int right(int[] nums) { int start = 0; int end = nums.length - 1; int target = 0; while (start target) { end = mid - 1; } else { start = mid + 1; } } return end; } Did this after the second hint -> Found left most of Zero and One and solved.: public static int maximumCount2(int[] nums) { int l = left(nums, 0); int r = left(nums, 1); return Math.max(l, nums.length - r); } private static int left(int[] nums, int target) { int start = 0; int end = nums.length - 1; while (start = target) { end = mid - 1; } else { start = mid + 1; } } return start; }

[@ivsc]

tried and colved it macha class Solution { public int maximumCount(int[] nums) { int c1=bsn(nums,0); int c2=bsn(nums,1); System.out.println(c1+" "+c2); return Math.max(c1, nums.length-c2); } public int bsn(int[] nums,int num) { int low=0,high=nums.length-1; while(low=num) { high=mid-1; } else low=mid+1; } return high+1; } }

[@Naveen_Reddy17]

Dont stop till we get enough 😅

[@saitejamukkerla]

2nd attempt left most of 1 and right most of -1 chesa class Solution { public int maximumCount(int[] arr) { int pos=Bsl(arr,1); int neg=Bsr(arr,-1); int positive=arr.length-pos; int negitive=neg+1; return Math.max(positive,negitive); } public static int Bsl(int []arr,int target){ int low=0; int high=arr.length-1; while(low=target){ high=mid-1; } else{ low=mid+1; } } if(low

[@rikkiteja4317]

Enti macha channel pettina starting loni break lu vundavu ani cheppav but intha valuable content neripisthu break lu isthunav maa lanti vallani andarni disappoint chesthunnav

[@engineeringanimuthyam]

@@rikkiteja4317 video cmng macha eroju kuda uploading lo undhi 🤧

[@edevenkatarajesh2622]

Brute force approach Time complexity is : O(n) Space complexity is : O(1) class Solution { public int maximumCount(int[] nums) { int n = nums.length; int neg = 0; int pos = 0; for (int i = 0; i < n; i++) { if (nums[i] < 0) { neg++; } else if (nums[i] > 0) { pos++; } } return Math.max(neg, pos); } } 💫💫💫💫

[@WORK-HARD-jm4cg]

solution chudakunda class Solution(object): def maximumCount(self, nums): count1=0 count2=0 for i in range(len(nums)): if nums[i] >0: count1=count1+1 elif nums[i]

[@Ammuuuammu-jf7hn]

Hii bro tcs code vita gurchi kodha chepandii bro ela preape avali ani plss ❤

[@WORK-HARD-jm4cg]

class Solution(object): def maximumCount(self, nums): x=self.countpositive(nums) y=self.countnegative(nums) print(y) return max(x,y) def countpositive(self,nums): l=0 r= len(nums)-1 while l0 and(mid==0 or nums[mid-1]

[@satyateja-ls7pk]

solved it in 1st attempt macha❤ class Solution { public int lzero(int[] nums){ int l =0; int ansl=0; int r = nums.length-1; int target=0; while(l=target){ r = mid-1; } else l = mid+1; } return l; } public int rzero(int[] nums){ int l =0; int ansr=0; int r = nums.length-1; int target=0; while(l

[@Bhargav3207]

macha without any helps and hints try chesa used left most and right most of 0 to get the answer class Solution { public int left(int[] nums){ int l=0; int r=nums.length-1; while(l

[@sathvikchintalapudi496]

macchhaa andharu edho comments peduthunte lite theeskunna kaani ee video lo nuv just chinna hint ivvagaane bale solution vacchindhi macchaa nijam ga edho teliyani excitement plzz macchaa ee series matram nuv asalu aapaku

[@engineeringanimuthyam]

🎊

[@jaswanthandhavarapu4325]

class Solution { public int maximumCount(int[] nums) { int countP = firstAndLastP(nums,false) - firstAndLastP(nums,true); int countN = firstAndLastN(nums,false) - firstAndLastN(nums,true); int max = Math.max(countP,countN); return max; } public int firstAndLastN(int[] nums, boolean findFirst){ int low = 0; int high = nums.length - 1; while(low

[@WORK-HARD-jm4cg]

Sorry to ask you again and again . But thappatle macha . Please em anukoku . Estimated month ? For the completion of entire DSA ? Cheppu macha kastha . Memu oka preparation lo untam

[@jaswanthandhavarapu4325]

bro naku telisi oka 4 - 5 months padutundhi because he is explaining all types of problems which we will be asked in online assements and online interviews

[@engineeringanimuthyam]

Inko 2 months lo basics ante under 10 lpa lopu adigevi finish cheyali next 2 months lo motham dsa finish cheyali

[@lavan1119]

class Solution { public static int pos(int[] nums) { int l = 0; int r = nums.length - 1; while (l 0) { l = mid; if (mid > 0 && nums[mid - 1] > 0) { l = mid - 1; } else { return r - l + 1; } } else { r = mid - 1; } } return 0; } public int neg(int[] nums) { int l = 0; int r = nums.length - 1; while (l

[@lavan1119]

macha time limit exceed