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Articles by Sunidhi Bansal
Page 24 of 81
Count of words that are present in all the given sentences in C++
We are given multiple sentences in the form of strings. The goal is to count the number of words that exist in all of the sentences.Note − words containing all lowercase letters will be considered onlyIf sentences are −“ I am learning C language ”“ learning new things is easy ““ Kids are learning healthy habits “Only “learning” exists in all three. So count is 1.Let us understand with examplesInput − “The clothes were dry”, “All the kids were playing”, “Those were the best days”Output − Count of words that are present in all the given sentences are − 2Explanation ...
Read MoreCount number of trailing zeros in (1^1)*(2^2)*(3^3)*(4^4)*.. in C++
Given an integer num as input. The goal is to find the number of trailing zeroes in the product 11 X 22 X 33 X…X numnum.For ExampleInputnum=5OutputCount of number of trailing zeros in (1^1)*(2^2)*(3^3)*(4^4)*.. are: 5ExplanationThe number of 2s and 5s in the product will be: 11 * 22* 33* 44* 55=11 * 22* 33* (22)4* 55. So total 10 2s and 5 5s, minimum is 5 so trailing zeroes will be 5.Inputnum=10OutputCount of number of trailing zeros in (1^1)*(2^2)*(3^3)*(4^4)*.. are: 5ExplanationThe number of 2s and 5s in the product will be: 11 *22*33*44*55*66 *77*88*99*1010 = 11 *22*33*44*55*66 *77*88*99*(2*5)10. So ...
Read MoreCount of Numbers such that difference between the number and sum of its digits not less than L in C++
We are given a number N and another number L. The goal is to find the numbers between 1 and N that have a difference between the number itself and the sum of its digits is not less than L.If N=23, L=10 then the count of such numbers will be 4.23-(2+3)=18, 22-(2+2)=18, 21-(2+1)=18, 20-(2+0)=18.All above numbers meet the conditionBut 19-(1+9)=9 which is less than L, similarly 18, 17….1.Let us understand with examplesInput − N=30 L=19Output − Count of Numbers such that difference between the number and sum of its digits not less than L are − 1Explanation − Only 30 ...
Read MoreCount number of trailing zeros in Binary representation of a number using Bitset in C++
Given an integer num as input. The goal is to find the number of trailing zeroes in the binary representation of num using bitset.A bitset stores the bits 0s and 1s in it. It is an array of bits.For ExampleInputnum = 10OutputCount of number of trailing zeros in Binary representation of a number using Bitset are: 1ExplanationThe number 10 in binary is represented as 1010 so trailing zeroes in it is 1.Inputnum = 64OutputCount of number of trailing zeros in Binary representation of a number using Bitset are: 6ExplanationThe number 64 in binary is represented as 10000000 so trailing zeroes ...
Read MoreCount of obtuse angles in a circle with 'k' equidistant points between 2 given points in C++
We are given a circle with K equidistant points on its circumference. Also we are given two points A and B. The goal is to count the number of triangles possible using these points such that they have an obtuse angle ACB( angle greater than 90o) inside them. The points A and B are such that A < B.Here K=8, A=2, B=5, count of points=2 (C, C’) such that angle LACB, LAC’B are obtuse.Let us understand with examplesInput − k=10, A=2, B=4Output − Count of obtuse angles in a circle with ‘k' equidistant points between 2 given points are − ...
Read MoreCount number of ways to jump to reach end in C++
Given an array of positive numbers. Each element represents the maximum number of jumps that can be made from that index to reach the end of the array. The goal is to find the number of jumps that can be made from that element to reach the end. If arr[] is [ 1, 2, 3 ] then for 1 jumps can be 1, for 2 jumps can be 1 or 2 and for 3 jumps can be made 1, 2 or 3.For ExampleInputarr[] = {1, 2, 3}OutputCount of number of ways to jump to reach end are: 1 1 0ExplanationFor ...
Read MoreCount of occurrences of a "1(0+)1" pattern in a string in C++
We are given a string str containing 0s, 1s and other alphabets . It also contains patterns of the form “1(0+)1” where 0+ means any number (>0) of consecutive 0s. The goal is to find such patterns ( “1(0+)1” ) inside string str.Let us understand with examplesInput − str = “abb010bb10111011”Output − Count of occurrences of a “1(0+)1” pattern in a string are − 2Explanation − The patterns inside str are highlighted: “abb010bb10111011”, “abb010bb10111011”Input − str = “01001011001001100”Output − Count of occurrences of a “1(0+)1” pattern in a string are − 4Explanation − The patterns inside str are highlighted: “01001011001001100”, ...
Read MoreCount number of ways to partition a set into k subsets in C++
Given two numbers e and p. The goal is to count the number of ways in which we can divide e elements of a set into p partitions/subsets.For ExampleInpute=4 p=2OutputCount of number of ways to partition a set into k subsets are: 7ExplanationIf elements are: a b c d then ways to divide them into 2 partitions are: (a, b, c)−(d), (a, b)−(c, d), (a, b, c)−(d), (a)−(b, c, d), (a, c)−(b, d), (a, c, d)−(b), (a, b, d)−(c). Total 7 ways.Inpute=2 p=2OutputCount of number of ways to partition a set into k subsets are: 1ExplanationIf elements are: a b ...
Read MoreCount of pairs of (i, j) such that ((n % i) % j) % n is maximized in C++
We are given a number num as input. The goal is to find the number of pairs of form (i, j) such that ((num%i)%j)%num is maximized and i and j both are in range [1, num].Let us understand with examplesInput − num=4Output − Count of pairs of (i, j) such that ((n % i) % j) % n is maximized are − 3Explanation − Pairs will be: (3, 2), (3, 3), (3, 4)Input − num=6Output − Count of pairs of (i, j) such that ((n % i) % j) % n is maximized are − 4Explanation − Pairs will be: ...
Read MoreCount numbers < = N whose difference with the count of primes upto them is > = K in C++
Given two integers N and K, the goal is to find the count of numbers such that they follow below conditions −Number=K Where count is the number of prime numbers less than or equal to Number.For ExampleInputN = 5, K = 2OutputCount of numbers < = N whose difference with the count of primes upto them is > = K are: 2ExplanationThe numbers that follow the conditions are: 5 ( 5−2>=2 ) and 4 ( 4−2>=2 )InputN = 10, K = 6OutputCount of numbers < = N whose difference with the count of primes upto them is > = K ...
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