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

Suppose we have a binary string s, initially this is say “0”. Now in each iteration invert it, and append it, thus after nth iteration, we will find the kth bit. Suppose the number of iteration is 4, and k = 7, so it will be −IterationValue (Initially 0)1012011030110100140110100110010110so 7th bit is 1.In each iteration, find complement, and append, thus after nth iteration, finds kth bitExample Live Demo#include using namespace std; string getComplement(string bin){    string temp = "";    for(int i= 0; i

Consider we have a number n, we have to find the sum of even indexed binomial coefficients like $$\left(\begin{array}{c}n\ 0\end{array}\right)+\left(\begin{array}{c}n\ 2\end{array}\right)+\left(\begin{array}{c}n\ 4\end{array}\right)+\left(\begin{array}{c}n\ 6\end{array}\right)+...\left(\begin{array}{c}4\ 0\end{array}\right)+\left(\begin{array}{c}4\ 2\end{array}\right)+\left(\begin{array}{c}4\ 4\end{array}\right)++=1+6+1=8$$So here we will find all the binomial coefficients, then only find the sum of even indexed values.Example Live Demo#include using namespace std; int evenIndexedTermSum(int n) {    int coeff[n + 1][n + 1];    for (int i = 0; i Read More

Consider we have an expression with brackets. If the index of one starting bracket is given, we have to find the closing ending bracket of that. So if the expression is like: (25*6+(88-32+(50/10)+20)), and the index of opening bracket is 6, then closing bracket will be at position 23.Here we will use the stack data-structure to solve this problem. We will traverse the expression from given index, and start pushing the opening brackets, when closing bracket is found, then pop element from stack, when the stack is empty, then return the index.Example Live Demo#include #include using namespace std; void getEndingBracketIndex(string exp, ... Read More

Suppose we have a positive number k. We have to find the positive number n, such that XOR of n and n+1 is same as k. So if k = 7 (111), output will be 3. As 3 (011), and 3 + 1 = 4 (100), so 011 XOR 100 = 111 (7)There are two possible cases. Consider n is even. The last bit of n = 0. Then the last bit of n + 1 = 1. Rest of the bits are same. So XOR will be 1, when n is odd, last bit 1, and last bit of ... Read More

Suppose we have a binary string bin. Then apply n iterations on it, and in each iteration 0 becomes 01 and 1 becomes 10, after that ith index character in the string after nth iteration. So if the binary string is 101, and n = 2, and i = 3, so after first iteration it will be 100110, in the next iteration, it will be 100101101001, so ith index is holding 1.To solve this, we have to follow these steps −Run loop n times, and in each iteration run another loop on the stringConvert each character of binary string, and ... Read More

Consider we have a point P in 2D plane and equation of a line, the task is to find the foot of the perpendicular from P to the line.The equation of the straight line is ax + by + c = 0. Equation of line passing through P and perpendicular to line. Equation of line passing through P and Q will be ay – bx + d = 0. Also P(x1, y1), and Q(x2, y2), so we put coordinate of P on the equation.ay 1−bx 1+d=0, so d=bx1−ay 1Also Q is the intersection of the given line and the line ... Read More