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We are given with a positive integer K and an array Ops[] which contains integers. The goal is to find the number of operations required to reduce K such that it becomes less than 0. Operations are −First operation is K + Ops[0], first element added to KAfter 1. Add Ops[i] to K until K
We are given a matrix of size NXN. The goal is to find the count of all valid pairs of indexes (i, j) such that the sum elements of column j is greater than the sum of elements of row i.We will do this by traversing the matrix and calculate sums of elements of each row and column.Store sum of elements of each in rowsum[N] and sum of elements of each column in colsum[N].Now make pairs of rowsum[i] and colsum[j] and check if colsum[j]>rowsum[i]. If true increment count for such pairs.Let’s understand with examples.Input-: matrix= { { 1, 2, ... Read More
We are given with an array arr[] of N elements. The goal is to find the count of all valid pairs of indexes (i, j) such that either arr[i] is divisible by arr[j] or arr[j] is divisible by arr[i] and i!=j.We will do this by traversing the array arr[] using two for loops for each number of pair and check if arr[i]%arr[j]==0 or arr[j]%arr[i]==0 when i!=j. If true increment count of pairs.Let’s understand with examples.Input − Arr[]= { 2, 4, 3, 6 } N=4Output − Count of valid pairs − 3Explanation − Valid pairs are −Arr[0] & Arr[1] → (2, ... Read More
We are given an array which contains the length of sides of triangles. The goal is to find the number of possible triangles that can be made by taking any three sides from that array.We will do this by checking if the sum of any two is always > third side. If yes these three sides can make a triangle. Increment count of possible triangles that can be made.Let’s understand with examples.Input − arr[]= {1, 2, 4, 5}Output − Count of possible triangles − 1Explanation − Sides (2, 4, 5) can only make a triangle as 2+4>5 & 4+5>2 & ... Read More
We are given two numbers N and K. The goal is to find the count of numbers between 1 and N that have divisors equal to the divisors of K in range [1, N].We will first count the divisors of K in range [1, N] and store in variable count.Now we will start from i=1 to i=N. Now for each number num=i (such that i!=K), count divisors of num in range[1, N]. And store their occurrences in variable divisors.If divisors=count means num has the same divisors as K in range [1, N]. increment count of such numbers.Let’s understand with examples.Input ... Read More
We are given with an array of distinct elements that are unsorted. The goal is to find the cross lines after the array is sorted. Cross lines are counted as shown below −Arr[]={ 1, 2, 4, 3, 5 } There are 3 cross lines as shown belowArr[]= { 1, 2, 3, 4, 5 }. There are no cross lines as the array is already sorted.We will count the cross lines using insertion sort in which an element from right is added to sorted elements on its left. Each time the element is added in the sorted part, increment count as ... Read More
We are given four integers L, R, A and B. The goal is to find the count of numbers in range [L, R] that fully divide either A or B or both.We will do this by traversing from L to R and for each number if number%A==0 or number%B==0 then increment count of divisors.Let’s understand with examples.Input − L=10, R=15, A=4, B=3Output − Count of divisors of A or B − 2Explanation −Number 12 is fully divisible by 3 and 4. Number 15 is fully divisible by 3 only. Total divisors=2Input − L=20, R=30, A=17, B=19Output − Count of divisors ... Read More
We are given with three arrays A[], B[] and C[]. The goal is to find all triplets of elements of these arrays such that A[i]
We are given with an array of numbers Arr[]. The goal is to count the number of pairs whose difference is equal to the maximum difference of all possible pairs. Count pairs (i!=j) and arr[x]- arr[y] is maximum possible.We will do this by first finding the maximum difference where (i!=j). And store as maxdiff. Then count all those pairs that have difference=maxdiff.Let’s understand with examples.Input − arr[]= { 1, 2, 3, 2, 4, 1, 5 }Output − No. of ways of choosing pair with maximum difference − 2Explanation −Here minimum no. is 1 and maximum number is 5, maximum difference ... Read More
We are given with an array of N elements. The goal is to find the count of all pairs (Arr[i], Arr[j]) which have a sum which is a perfect square such that i!=j. That is Arr[i]+Arr[j] is a perfect square.We will do this by calculating the sum of pairs and check if the square root of that sum is equal to the floor value of the square root. sqrt(Arr[i]+Arr[j])-floor( sqrt(Arr[i]+Arr[j] )==0.Let’s understand with examples.Input − Arr[]= { 4, 3, 2, 1, 2, 4 } N=6Output − Count of pairs with sum as perfect square − 2Explanation −Arr[1]+Arr[3]=4, sqrt(4)-floor(4)=0 4 is ... Read More