Find pairs in an array such that a+b%10 = k - c++

Question

There is a ordered list like
A=[7, 9, 10, 11, 12, 13, 20]
and I have to find pairs a+b%10=k where 0<=k<=9
For example k = 0
Pairs: (7, 13), (9, 11), (10, 20)
How can i find the number of pairs in O(n) time?
I tried to find convert all the list with take mod(10)
for (auto i : A) {
if (i <= k) {
B.push_back(i);
}
else {
B.push_back(i % 10);
}
}
After that i tried to define summations that gives k via unorderep_map
unordered_map<int, int> sumList;
int j = k;
for (int i = 0; i < 10; i++) {
sumList[i] = j;
if (j==0) j=9;
j--;
}
But i can't figure out that how can i count the number of pairs in O(n), what can i do now?

Answer 1

Let’s begin with a simple example. Assume that k = 0. That means that we want to find the number of pairs that sum up to a multiple of 10. What would those pairs look like? Well, they could be formed by
adding up a number whose last digit is 1 with a number whose last digit is 9,
adding up a number whose last digit is 2 with a number whose last digit is 8,
adding up a number whose last digit is 3 with a number whose last digit is 7,
adding up a number whose last digit is 4 with a number whose last digit is 6, or
adding up two numbers whose last digit is 5, or
adding up two numbers whose last digit is 0.
So suppose you have a frequency table A where A[i] is the number of numbers with last digit i. Then the number of pairs of numbers whose last digits are i and j, respectively, is given by
A[i] * A[j] if i ≠ j, and
A[i] * A[i-1] / 2 if i = j.
Based on this, if you wanted to count the number of pairs summing to k mod 10, you could
fill in the A array, then
iterate over all possible pairs that sum to k, using the above formula to count up the number of pairs without explicitly listing all of them.
That last step takes time O(1), since there are only ten buckets and iterating over the pairs you need therefore requires at most a constant amount of work.
I’ll leave the rest of the details to you.
Hope this helps!

Answer 2

You can modify counting sort for this.
Below is an untested, unoptimized and only illustrative version:
int mods[10];
void count_mods(int nums[], int n) {
for (int i = 0; i < n; i++)
mods[nums[i]%10]++;
}
int count_pairs(int k) {
// TODO: there's definitely a better way to do this, but it's O(1) anyway..
int count = 0;
for (int i = 0; i < 10; i++)
for (int j = i+1; j < n; j++)
if ((i + j) % 10 == k) {
int pairs = mods[i] > mods[j] ? mods[j] : mods[i];
if (i == j)
pairs /= 2;
count += pairs;
}
return count;
}
EDIT:
With a smaller constant.
int mods[10];
void count_mods(int nums[], int n) {
for (int i = 0; i < n; i++)
mods[nums[i]%10]++;
}
int count_pairs(int k) {
int count = 0;
for (int i = 0; i < 10; i++) {
int j = k - i;
if (j < 0)
j += 10;
count += min(mods[i], mods[j]);
// When k = 2*i we count half (rounded down) the items to make the pairs.
// Thus, we substract the extra elements by rounding up the half.
if (i == j)
count -= (mods[i]+1) / 2;
}
// We counted everything twice.
return count / 2;
}