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find max occurance of name from list annotation for calling kotlin class/function in java why kotlin has primary and secondary constructor async await multiple call inside launch async await vs withContext coroutine builder implementation of pagination data source in pagination navigation component safe args custom view api vs implementation glide internal working solid principle in java @Bind vs @Provides in DI code coverage room database downgrade
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Senior Android Developer

Interviewed at Infogain

3.6★
Dec 6, 2023

find max occurance of name from list annotation for calling kotlin class/function in java why kotlin has primary and secondary constructor async await multiple call inside launch async await vs withContext coroutine builder implementation of pagination data source in pagination navigation component safe args custom view api vs implementation glide internal working solid principle in java @Bind vs @Provides in DI code coverage room database downgrade

1- we are given two strings S and T consisting of N and M characters , respectively, and you would like to check whether they might have been obtained as OCR scans of the same text. for example, both strings "A2Le" and "2pl1" could have been obtained as scans of the word "Apple" (but also as scans of the word "Ample"). Both strings "a10" and "10a" could have been obtained as scans of the word "abbbbbbbbba" (but also from many other strings of length 11, starting and ending with "a") . on the other hand, string "ba1" and "1Ad" could not have been obtained from the same text since the second letter of each text is different. so for "A2Le" and "2pL1" we return true. for "a10" and "10a" we return true. for "ba1" and "1Ad" we return false. for "3x2x" and "8" we return false. 2- we have a function like this in java : public String solution(String S) { } String S is consisting of letters A, B, C, D. the string can be transformed either by removing a letter A together with an adjacent letter B , or by removing letter C together with an adjacent letter D . the function should return any string that can be obtained from S by repeatedly applying the described transformation to the point that cannot be further transformed. example 1 : given "CBACD" the function may return "C" . because one of the possible sequence would be : "CBACD" -> "CBA" -> "C" example 2: given "CABABD" the function may return an empty string because one of the possible sequence would be : "CABABD" -> "CABD" -> "CD" -> "" example 3 : "ACBDACBD" the function returns "ACBDACBD" because no operation can be applied to string S . the length of string is within the range of [0 .. 250000] 3- we have a function like this in kotlin : fun solution(A: IntArray, B: IntArray, X: Int, Y: Int): Int { } A and B are intArrays contain values within 1 to 100. values of A and B are coordinations like this (x,y) and they have the same length. the X and Y make one coordination like (X,Y). now if (X,Y) are in the range of +20 or -20 of any of the values in A and B return the index of that pair. example 1 : A = [100, 200, 100] , B = [50,100,100] , X=100 , Y = 70 the answer is 0 . example 2: A = [100,200,100] , B = [50,100,100] , X=100 , Y = 100 the answer is 2 .
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Senior Android Developer

Interviewed at Zalando

3.6★
Dec 17, 2023

1- we are given two strings S and T consisting of N and M characters , respectively, and you would like to check whether they might have been obtained as OCR scans of the same text. for example, both strings "A2Le" and "2pl1" could have been obtained as scans of the word "Apple" (but also as scans of the word "Ample"). Both strings "a10" and "10a" could have been obtained as scans of the word "abbbbbbbbba" (but also from many other strings of length 11, starting and ending with "a") . on the other hand, string "ba1" and "1Ad" could not have been obtained from the same text since the second letter of each text is different. so for "A2Le" and "2pL1" we return true. for "a10" and "10a" we return true. for "ba1" and "1Ad" we return false. for "3x2x" and "8" we return false. 2- we have a function like this in java : public String solution(String S) { } String S is consisting of letters A, B, C, D. the string can be transformed either by removing a letter A together with an adjacent letter B , or by removing letter C together with an adjacent letter D . the function should return any string that can be obtained from S by repeatedly applying the described transformation to the point that cannot be further transformed. example 1 : given "CBACD" the function may return "C" . because one of the possible sequence would be : "CBACD" -> "CBA" -> "C" example 2: given "CABABD" the function may return an empty string because one of the possible sequence would be : "CABABD" -> "CABD" -> "CD" -> "" example 3 : "ACBDACBD" the function returns "ACBDACBD" because no operation can be applied to string S . the length of string is within the range of [0 .. 250000] 3- we have a function like this in kotlin : fun solution(A: IntArray, B: IntArray, X: Int, Y: Int): Int { } A and B are intArrays contain values within 1 to 100. values of A and B are coordinations like this (x,y) and they have the same length. the X and Y make one coordination like (X,Y). now if (X,Y) are in the range of +20 or -20 of any of the values in A and B return the index of that pair. example 1 : A = [100, 200, 100] , B = [50,100,100] , X=100 , Y = 70 the answer is 0 . example 2: A = [100,200,100] , B = [50,100,100] , X=100 , Y = 100 the answer is 2 .

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