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Articles by Sunidhi Bansal
Page 24 of 81
Count 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 ...
Read MoreCount of strings where adjacent characters are of difference one in C++
We are given a num number as input. The goal is to count the number of possible strings of length num such that all adjacent characters have difference between ascii values as 1.If num is 2 then strings will be “ab”, “ba”, “bc”, “cb”, ……..”yz”, “zy”.Let us understand with examplesInput − num=3Output − Count of strings where adjacent characters are of difference one are − 98Explanation − Some sample strings are: “abc”, “aba”, “cde” …..”xyx”, “zyz”, “xyz”.Input − num=2Output − Count of strings where adjacent characters are of difference one are − 50Explanation − Some sample strings are: “ab”, “ba”, ...
Read MoreCount the number of pop operations on stack to get each element of the array in C++
Given an array of numbers and a stack. All the elements of the array are present inside the stack The goal is to find the count of pop operations required for getting individual array elements.The stack is filled in decreasing order, the first element is highest and top element is lowest.For ExampleInputStack [ 7, 6, 2, 1 ] array : 2, 1, 6, 7OutputCount of number of pop operations on stack to get each element of the array are: 3 1 0 0ExplanationTraversing array from 0th index, To get 2 we will pop stack three times. So arr[0] is 3. ...
Read MoreCount of distinct rectangles inscribed in an equilateral triangle in C++
We are an equilateral triangle with side length. The goal is to count the number of distinct rectangles that can be present inside the triangle such that horizontal sides of the rectangle are parallel to the base. Also all end points of the rectangle touch the dots as shown.Let us understand with examplesInput − sides=3Output − Count of distinct rectangles inscribed in an equilateral triangle are − 1Explanation − The figure above shows the rectangle.Input − sides=10Output − Count of distinct rectangles inscribed in an equilateral triangle are − 200Approach used in the below program is as followsAs from the ...
Read MoreCount of all possible values of X such that A % X = B in C++
Given two integers A and B and a number X. The goal is to find the count of values that X can have so that A%X=B. For the above equation if, A==B then infinite values of X are possible, so return −1. If A < B then there would be no solution so return 0. If A>B then return count of divisors of (AB) as result.For ExampleInputA=5, B=2OutputCount of all possible values of X such that A % X = B are: 1Explanation5%3=2. So X is 3 here.InputA=10, B=10OutputCount of all possible values of X such that A % X ...
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