Suppose we have a number N. We have to find minimum number that divide N to make it perfect square. So if N = 50, then minimum number is 2, as 50 / 2 = 25, and 25 is a perfect square.A number is perfect square if it has even number of different factors. So we will try to find the prime factors of N, and find each prime factor power. Find and multiply all prime factors whose power is odd.Example Live Demo#include #include using namespace std; int findMinimumNumberToDivide(int n) {    int prime_factor_count = 0, min_divisor = 1;    while ... Read More

Suppose there is a circle, and there are n petrol pumps on the circle. We have two sets of data like −The amount of petrol that every petrol pump hasDistance from one petrol pump to another.Calculate the first point, from where a truck will be able to complete the circle. Assume for 1 liter petrol, the truck can go 1 unit of distance. Suppose there are four petrol pumps, and the amount of petrol, and distance from the next petrol pump is as like [(4, 6), (6, 5), (7, 3), (4, 5)], the first point from where truck can make ... Read More

As we know that log(x*y) = log(x) + log(y). So we will see what are the minimum number of log values are required to calculate all log values from 1 to N. So if N is 6, then output will be 3, as from log(1) to log(6), there are three log values are required except log(1). As log(1) is always 0, then ignore it, now for log(2) and log(3), we have to find. After that for log(4) this is log(2)+ log(2), but value of log(2) is known, so we do not calculate this again, for log(5), we need to calculate. ... Read More

Suppose we have a number position in infinite number line. (-inf to +inf). Starting from 0, we have to reach to the target by moving as described. In ith move, we can go i steps either left or right. We have to find minimum number of moves that are required. Suppose the target is 2, so minimum steps will be 3. From 0 to 1, from 1 to -1 and from -1 to 2.To solve this problem, we have some important points to remember. If the target is negative, then just take this as positive, as the number line is ... Read More

Suppose we have the area A and a perimeter P, now we have to find what will be the maximum volume that can be made in form of cuboid from given perimeter and surface area. So when the P is 24 and A is 24, then the output will be 8.As we know for given perimeter of cuboid P = 4(length + breadth + Depth), for area, it will be A = 2(length* breadth + breadth*Depth + length *Depth), and the volume is V = (length* breadth*Depth)Example Live Demo#include #include using namespace std; float maxVolumeCuboid(float Peri, float Area) {    float ... Read More

Suppose there are N stations in the straight line. Each of them has same non-negative power of radiation power. Every station can increase the radiation power of its neighboring stations in the following way.Suppose the station i with radiation power R, will increase (i – 1)th station’s radiation power, by R-1, (i - 2)th station’s radiation power by R-2, and will increase (i + 1)th station’s radiation power, by R-1, (i + 2)th station’s radiation power by R-2. So on. So for example, if the array is like Arr = [1, 2, 3], then the output will be 3, 4, ... Read More

Suppose we have an array of n integers. Find the max sum of strictly increasing subarrays. So if the array is like [1, 2, 3, 2, 5, 1, 7], the sum is 8. In this array there are three strictly increasing sub-arrays these are {1, 2, 3}, {2, 5} and {1, 7}. The max sum sub-array is {1, 7}To solve this problem, we have to keep track of max sum and the current sum. For each element arr[i] if this is larger than arr[i – 1], then we add this to the current sum, otherwise arr[i] is the starting point ... Read More

Suppose we have an array of n elements. Find the maximum possible sum of all elements such that all the elements are same. Only operation that is allowed is choosing any two elements and replacing the larger of them by the absolute difference of the two. Suppose elements are like [9, 12, 3, 6]. Then the output will be 12. So at first replace A[1] with A[1] – A[3] = 12 – 6 = 6. So now elements are [9, 6, 3, 6], then replace A[3] with A[3] – A[2] = 6 – 3 = 3. So the elements are ... Read More

Suppose we have a number n. We have to find the length of period in decimal value of 1/n. So if the value of n is 7, then 1/7 = 0.142857142857… That part in bold letters are repeating. So here the length of period is 6.For a number n, there can be n distinct remainders in the output, but the period may not begin from the first remainder as some initial remainders are non-repeating. So we have to make sure that a remainder from period is picked, start from (n+1)th remainder, and start looking for the next occurrence. The distance ... Read More

Consider we have an array A with few elements. We have to find sum of all distinct elements in the array. So if the A = [5, 12, 63, 5, 33, 47, 12, 63], then sum of distinct elements is 160. Duplicate elements are simply ignored once it has considered.We can use the unordered set to solve this problem efficiently. We will run one single for loop and which value comes first time, its add in sum variable and store in hash table that for next time, we will not use this value.Example Live Demo#include #include using namespace std; int getNonRepeatSum(int ... Read More