Programming Articles - Page 936 of 3366

Suppose we have limited coins of denominations (₹1, ₹2, ₹5 and ₹10). We have to find in how many ways can you sum them up to a total of ₹n? We have an array count of size 4, where count[0] indicates coins of ₹1, count[1] indicates coins of ₹2 and so on.So, if the input is like n = 25 count = [7, 3, 2, 2], then the output will be 9.To solve this, we will follow these steps −denom := [1, 2, 5, 10]A := an array of size (n + 1) and fill with 0B := a new ... Read More

Suppose we have to calculate nCr values many times. We can solve this very efficient way. If we store the lower values of nCr we can easily find higher values. So if we have n, we have to find list of nC0 to nCn. If answer is too large then return that modulo 10^9.So, if the input is like n = 6, then the output will be [1, 6, 15, 20, 15, 6, 1].To solve this, we will follow these steps −items := a list with single element 1for r in range 1 to n, doinsert floor of (last element ... Read More

Suppose we want to find the total area covered by two rectilinear rectangles in a 2D plane. Here each rectangle is defined by its bottom left corner and top right corner as shown in the figure.To solve this, we will follow these steps −width_1 := |C-A|, height_1 := |D-B|width_2 := |G-E|, height_2 := |H-F|area := width_1*height_1 + width_2*height_2if (GC) or (F>D) or (HD) or (H

Suppose we have a sequence nums of size n. We have to find the maximum size of subsequence of nums in which every pair (p, q) is a nice pair? A pait is said to be nice pair if and only if it holds at least one of these conditions: 1. The parity of the number of distinct prime divisors of p is equal to that of b. For example, the value 18 has two distinct prime divisors: 2 and 3. 2. The parity of the sum of all positive divisors of p is same as q.So, if the input ... Read More

Suppose we have a number A. We have to generate a large number X by concatenating A, n times in a row and find the value of X modulo m.So, if the input is like A = 15 n = 3 m = 8, then the output will be 3, because the number x will be 151515, and 151515 mod 8 = 3.To solve this, we will follow these steps −if A is same as 0, thenreturn 0an:= Ac:= number of digits in Ac:= 10^cd:= c-1newmod := d*mval := (c ^ n mod newmod) -1val :=(val + newmod) mod newmodan ... Read More

Suppose there is a strange language called Ajob language. It has infinite number of letters. We know n words in this language. The first word is one character long, second one is two character long and so on. And all letters in a word are unique. If we select any one of the n words and form a subsequence from it. The length of the subsequence should be k less than the length of original word. For an example, if the length of the chosen word is say L, then the length of the subsequence should be (L - k). ... Read More

Suppose we have a number n, and another number k. We have to check whether n can be represented as a sum of k prime numbers or not.So, if the input is like n = 30 k = 3, then the output will be True because 30 can be represented like 2 + 11 + 17.To solve this, we will follow these steps −if n < k*2, thenreturn Falseif k > 2, thenreturn Trueif k is same as 2, thenif n is even, thenreturn Trueif (n-2) is prime, thenreturn Truereturn Falseif n is prime, thenreturn Truereturn FalseExampleLet us see the ... Read More

Suppose there is a function f(x), that counts number of (p, q) pairs, such that1 < p

Suppose we have a number n. If we have numbers like [1, 2, ..., n] we have to count number ofpossible BSTs can be formed using these n values. If the answer is too large, then mod the result by 10^9+7.So, if the input is like n = 3, then the output will be 14, To solve this, we will follow these stepsa := a list with values [0, 1]m := 10^9+7max_n := 1000for k in range 2 to max_n + 1, doinsert (1 + sum of all elements of the list (a[i] * a[k - i] for all i ... Read More

Suppose, there are n number of balls. The balls are ordered in a fashion 1, 2, 3, 4, ..., n. Now the balls are reversed in order, or ordered in a way n, n-1, n-2, ......, 2, 1. The balls are again reversed in order, this time they are reversed from position 1 to n, or now the order becomes n, 1, 2, ....., n-1. This reversing process is repeated n times and each time the starting position is moved 1 place to the right. We now have to find out the position of a ball initially in position 'index' ... Read More