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A7
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A7
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Q1 - Given an integer array(arr) and its size(n), print the count of odd and even integers present in the array.
Solution:
#include <iostream>
using namespace std;
int main(){
int n;
cout<<"Enter size of array: ";
cin>>n;
int arr[n];
cout<<"Enter elements in an array: ";
for(int i=0;i<n;i++){
cin>>arr[i];
}
int c1=0,c2=0;
for(int i=0;i<n;i++){
if(arr[i]%2==0){
c1++;
}
else{
c2++;
}
}
cout<<"Even Numbers: "<<c1<<endl;
cout<<"Odd Numbers: "<<c2<<endl;
return 0;
}
---------------------------------------------------------------------------------------
Q2 - Given an integer array and its size, find the sum of the greatest and the smallest integer present
in the array. Here 1< size <101
Solution:
#include <iostream>
using namespace std;
int main(){
int n;
cout<<"Enter size of array: ";
cin>>n;
int arr[n];
cout<<"Enter elements in an array: ";
for(int i=0;i<n;i++){
cin>>arr[i];
}
int min=arr[0],max=arr[0];
for(int i=0;i<n;i++){
if(arr[i]>max){
max=arr[i];
}
if(arr[i]<min){
min=arr[i];
}
}
cout<<"Sum of smallest and largest element in an array is: "<<min+max;
return 0;
}
---------------------------------------------------------------------------------------
Q3 - Given an integer array and its size, reverse the array and print it . Here 1<size<101
Solution:
#include <iostream>
using namespace std;
int main(){
int n;
cout<<"Enter size of array: ";
cin>>n;
int arr[n];
cout<<"Enter elements in an array: ";
for(int i=0;i<n;i++){
cin>>arr[i];
}
cout<<"reverse of an array is: ";
for(int i=n-1;i>=0;i--){
cout<<arr[i]<<" ";
}
return 0;
}
----------------------------------------------------------------------------------------
Q4 - Given two arrays a[] and b[] of same size.Your task is to find the minimum sum of two elements such
that they belong to different arrays and are not at the same index. Here 1<size<101
Solution:
#include<iostream>
using namespace std;
int main(){
int n;
cout<<"Enter number of elements: "<<endl;
cin>>n;
int a[n],b[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
for(int i=0;i<n;i++){
cin>>b[i];
}
int min1=a[0],min2=b[0];
for(int i=0;i<n;i++){
if(a[i]<min1){
min1=a[i];
}
if(b[i]<min2){
min2=b[i];
}
}
cout<<"Minimum sum of two array: "<<min1+min2<<endl;;
}
-----------------------------------------------------------------------------------------
Q5 - Given an array containing n distinct integers in the range [0,n] (inclusive of both 0 and n) (inclusive of
both 0 and n). Find and return the only number of the range that is not present in the array. Here 1<n<101.
Solution:
#include<iostream>
using namespace std;
int main(){
int n;
cout<<"Enter number of elements: ";
cin>>n;
int arr[n],missing=-1,summ=0,sum=0;
for(int i=0;i<n;i++){
cin>>arr[i];
}
summ=n*(n+1)/2;
for(int i=0;i<n;i++){
sum+=arr[i];
}
missing=summ-sum;
cout<<"Missing number is "<<missing;
}
------------------------------------------------------------------------------------------
Q6 - Given an integer array containing n elements. Find the element in the array for which all the elements
to its left are smaller than it and all the elements to the right of it are larger than it. Here 1<n<101.
Solution:
#include<iostream>
using namespace std;
int main(){
int n;
cout<<"Enter number of elements: ";
cin>>n;
int arr[n];
for(int i=0;i<n;i++){
cin>>arr[i];
}
int max[n],min[n],element=-1,big=arr[0],small=arr[n-1];
for(int i=0;i<n;i++){
if(arr[i]>big){
big=arr[i];
}
max[i]=big;
}
for(int i=n-1;i>=0;i--){
if(arr[i]<small){
small=arr[i];
}
min[i]=small;
}
for(int i=0;i<n;i++){
if(i!=0 && i!=n-1){
if(min[i]==max[i]){
element=min[i];
break;
}
}
}
cout<<element;
}