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Missing even and odd elements from the given arrays in C++
Problem statement
Given two integer arrays even [] and odd [] which contains consecutive even and odd elements respectively with one element missing from each of the arrays. The task is to find the missing elements.
Example
If even[] = {10, 8, 6, 16, 12} and
odd[] = {3, 9, 13, 7, 11} then
missing number from even array is 14 and from odd array is 5.Algorithm
- Store the minimum and the maximum even elements from the even[] array in variables minEven and maxEven
- Sum of first N even numbers is N * (N + 1). Calculate sum of even numbers from 2 to minEven say sum1 and sum of even numbers from 2 to maxEven say sum2
- The required sum of the even array will be reqSum = sum2 – sum1 + minEven, subtracting the even[] array sum from this reqSum will give us the missing even number
- Similarly, the missing odd number can also be found as we know that the sum of first N odd numbers is N2
Example
#include <bits/stdc++.h>
using namespace std;
void findMissingNums(int even[], int sizeEven, int odd[], int sizeOdd) {
int minEven = INT_MAX;
int maxEven = INT_MIN;
int minOdd = INT_MAX;
int maxOdd = INT_MIN;
int sumEvenArr = 0, sumOddArr = 0;
for (int i = 0; i < sizeEven; i++) {
minEven = min(minEven, even[i]);
maxEven = max(maxEven, even[i]);
sumEvenArr += even[i];
}
for (int i = 0; i < sizeOdd; i++) {
minOdd = min(minOdd, odd[i]);
maxOdd = max(maxOdd, odd[i]);
sumOddArr += odd[i];
}
int totalTerms = 0, reqSum = 0;
totalTerms = minEven / 2;
int evenSumMin = totalTerms * (totalTerms + 1);
totalTerms = maxEven / 2;
int evenSumMax = totalTerms * (totalTerms + 1);
reqSum = evenSumMax - evenSumMin + minEven;
cout << "Missing even number = " << reqSum - sumEvenArr << "\n";
totalTerms = (minOdd / 2) + 1;
int oddSumMin = totalTerms * totalTerms;
totalTerms = (maxOdd / 2) + 1;
int oddSumMax = totalTerms * totalTerms;
reqSum = oddSumMax - oddSumMin + minOdd;
cout << "Missing odd number = " << reqSum - sumOddArr << "\n";
}
int main() {
int even[] = {10, 8, 6, 16, 12};
int sizeEven = sizeof(even) / sizeof(even[0]);
int odd[] = {3, 9, 13, 7, 11};
int sizeOdd = sizeof(odd) / sizeof(odd[0]);
findMissingNums(even, sizeEven, odd, sizeOdd);
return 0;
}When you compile and execute above program. It generates following output −
Output
Missing even number = 14 Missing odd number = 5
Updated on: 20-Dec-2019
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