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C++ Articles
Page 271 of 597
Count the number of non-increasing subarrays in C++
Given an array arr[] containing positive integers. The goal is to find the number of subarrays of length at least 1 which are non−increasing. If arr[]= {1, 3, 2}, then subarrays will be {1}, {2}, {3}, {3, 2}. Count is 4.For ExampleInputarr[] = {5, 4, 5}OutputCount of number of non-increasing subarrays are: 7ExplanationThe subarrays will be − {5}, {4}, {5}, {5, 4}Inputarr[] = {10, 9, 8, 7}OutputCount of number of non−increasing subarrays are − 10ExplanationThe subarrays will be − {10}, {9}, {8}, {7}, {10, 9}, {9, 8}, {8, 7}, {10, 9, 8}, {9, 8, 7}, {10, 9, 8, 7}Approach used ...
Read MoreCount the number of holes in an integer in C++
Given an array of holes[10] containing a number of holes in numbers 0 to 9. The goal is to find the number of holes in an integer number given as input. Given − holes[] = { 2, 1, 1, 0, 0, 1, 1, 1, 0 }For ExampleInputnumber = 239143OutputCount the number of holes in an integer are: 3ExplanationWe will count holes given in holes[] 239143 ( 1+0+0+2+0+0 )Inputnumber = 12345OutputCount the number of holes in an integer are: 3ExplanationWe will count holes given in holes[] 12345 ( 1+1+0+0+1)Approach used in the below program is as follows −Simply take each leftmost ...
Read MoreCount the number of intervals in which a given value lies in C++
Given a 2D array arr[][] containing intervals and a number ‘value’. The goal is to find the number of intervals present in arr between which value lies. For example intervals are [ [1, 5], [3, 7] ] and value=4 then it lies in both these intervals and count would be 2.For ExampleInputarr[4][2] = { { 1, 20 }, { 12, 25 }, { 32, 40 }, { 15, 18 } } value=16OutputCount of number of intervals in which a given value lies are: 3ExplanationThe value 16 lies between 1−20, 12−25 and 15−18Inputarr[4][2] = {{ 1, 20 }, { 20, 30 ...
Read MoreCount of Binary Digit numbers smaller than N in C++
Given an integer N as input. The goal is to find the count of integers that are less than N and represented in Binary form. For example if input N is 12 then numbers less than 12 are 1, 10, 11 that are binary and contain 0s and 1s as digits. The answer would be 3.For ExampleInputN=100OutputCount of Binary Digit numbers smaller than N are − 4ExplanationThe Binary numbers less than 100 are − 1, 10, 11, 100InputN=120OutputCount of Binary Digit numbers smaller than N are: 7ExplanationThe Binary numbers less than 100 are : 1, 10, 11, 100, 101, 110, ...
Read MoreCount of arrays in which all adjacent elements are such that one of them divide the another in C++
Given two integers named ‘one’ and ‘another’. The goal is to find the number of possible arrays such that −The elements in the array are in range between 1 and ‘another’.All elements of array are such that arr[i] divides arr[i+1] or arr[i+1] divides arr[i+2]....and so on.The length of the array is ‘one’.For ExampleInputone = 3, another = 2OutputCount of arrays in which all adjacent elements are such that one of them divide the another are: 8ExplanationThe arrays will be: [ 1, 1, 1 ], [ 1, 1, 2 ], [ 1, 2, 1 ], [ 1, 2, 2 ], [ ...
Read MoreCount of all N digit numbers such that num + Rev(num) = 10^N - 1 in C++
Given a number N as input. The goal is to find the count of all N digit numbers that have sum Count of all N digit numbers such that num + Rev(num) = 10N − 1num+rev(num)=10N−1For ExampleInputN=4OutputCount of all N digit numbers such that num + Rev(num) = 10N − 1 are − 90ExplanationThe numbers would be − 1. 1188 + 8811 = 9999 2. 2277 + 7722 = 9999 3. 1278 + 8721 = 9999 ……...total 90 numbersInputN=5OutputCount of all N digit numbers such that num + Rev(num) = 10N − 1 are − 0ExplanationAs N is odd, ...
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 ...
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 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 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 ...
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