Given with an array of int elements, the task is to arrange the elements in descending order and finding their occurrences.Input : arr[]={1, 1, 1, 2, 2, 2, 3, 3, 4, 5, 6, 7, 7} Output : 7 occurs: 2    6 occurs: 1    5 occurs: 1    4 occurs: 1    3 occurs: 2    2 occurs: 3    1 occurs: 3AlgorithmSTART Step 1 -> input array with elements in sorting order Step 2 -> calculate size of an array by sizeof(a)/sizeof(a[0] Step 3 -> store size in a variable say en Step 4 -> Loop For i=siz-1 ... Read More

In this article, we will learn how to print prime numbers from 1 to N in reverse order. A prime number is a number greater than 1 that has no positive divisors other than 1 and itself. Our goal is to find all prime numbers in the given range from 1 to N and then print them in reverse order. For example, given N = 20, the prime numbers between 1 and 20 are: 2, 3, 5, 7, 11, 13, 17, 19 The output should be the prime numbers printed in reverse order: 19, 17, 13, 11, 7, 5, ... Read More

Input N which is equivalent to the number till where series should be printedInput : N=5 Output : 0 ¼ ½ ¾ 1AlgorithmSTART Step 1 -> declare start variables as int num , den, i, n Step 2 -> input number in n Step 3 -> Loop For from i to 0 and i End For Loop STOPExample#include int main(int argc, char const *argv[]) {    int num, den;    int i, n;    printf("Enter series limit");    scanf("%d", &n); //Till where the series will be starting from zero    for (i = 0; i < n; i++) { ... Read More

Given two sorted arrays and output should display their uncommon elementsGiven : array1[]= {1, 4, 6, 9, 12}    array2[]= {2, 4, 7, 8, 9, 10} Output : 1 2 6 7 8 10 12AlgorithmSTART Step 1 -> declare two arrays array1 and array2 with elements as int and variables n1, n2, i to 0 and j to 0 Step 2 -> calculate number of elements in array1 sizeof(array1)/sizeof(array1[0]) Step 3-> calculate number of elements in array2 sizeof(array2)/sizeof(array2[0]) Step 4 -> Loop While till i loop While i < n1 && array1[i]!=array2[j]    Print array1[i++] Step 7 -> End Loop ... Read More

It will display the missing values from the given set entered by the userGiven : array = {88, 105, 3, 2, 200, 0, 10}; Output : 1 4-9 11-87 89-99AlgorithmSTART STEP 1-> Take an array with elements, bool flag[MAX] to Fale, int i, j, n to size of array Step 2-> Loop For from I to 0 and i=0       Set flag[array[i]]=true    End IF Step 3 -> End For Loop Step 4 -> Loop For from i to 0 and i=0) {          flag[array[i]] = true; //Making the value of the elements present in ... Read More

Input three strings and replace each string with a character which user has entered and then display edited strings. After that, concatenate edited strings and display them.Input:    string 1 : tutorials replacement character for string 1 : x    String 2 : points replacement character for string 2 : y    String 3: best replacement character for string 3 : z Output :    string 1: xxxxxxxxx    String 2 :yyyyyy    String 3 : zzzz    After concatenation : xxxxxxxxxyyyyyyzzzzAlgorithmSTART Step 1-> declare three array of characters str1, str2 and str3 with variables as ch1, ch2 and ch3 ... Read More

Input a string and find the total number of words, vowels and frequency of a character enter by a userInput : enter s string : I love my MOM      Enter a charcter of which you want to find a frequency: M    Total frequency of M : 2    Total number of vowels : 4    Total number of words : 4AlgorithmSTART Step 1 Declare array of string, ch, i, freq to 0, vow to 0, word to 0 Step 2 Input a string and a character ch Step 3 Loop for from i to 0 and str[i]!=’\o’ ... Read More

Given with n numbers program must find those n number whose sum is a perfect squareInput : 5 Output : 1 3 5 7 9 1+3+5+7+9=25 i.e (5)^2AlgorithmSTART    Step 1 : Declare a Macro for size let’s say of 5 and i to 1    Step 2: loop While till i printing (2*i)-1 Step       Step 2.2 -> incrementing i with 1 Step    Step3-> End loop While STOPExample#include # define SIZE 5 int main() {    int i=1;    while(i

In this section, we will see how we can get the sum of all odd prime factors of a number in an efficient way. There is a number say n = 1092, we have to get all factor of this. The prime factors of 1092 are 2, 2, 3, 7, 13. The sum of all odd factors is 3+7+13 = 23. To solve this problem, we have to follow this rule −When the number is divisible by 2, ignore that factor, and divide the number by 2 repeatedly.Now the number must be odd. Now starting from 3 to square root ... Read More

In this section, we will see how we can get the largest prime factor of a number in an efficient way. There is a number say n = 1092, we have to get the largest prime factor of this. The prime factors of 1092 are 2, 2, 3, 7, 13. So the largest is 13. To solve this problem, we have to follow this rule −When the number is divisible by 2, then store 2 as largest, and divide the number by 2 repeatedly.Now the number must be odd. Now starting from 3 to square root of the number, if ... Read More