C++ program to find last value of matrix with given constraints

Suppose we have a list of numbers A with N elements in it. The elements are 1, 2 or 3, Consider a number X[1][j] = A[j] where j is in range 1 to N. And X[i][j] = |X[i-1][j] - X[i-1][j+1]| where i is in range 2 to N and j is in range 1 to N+1-i. We have to find the value of X[i][j].

So, if the input is like A = [1,2,3,1], then the output will be 1, because

X[1][1] to X[1][4] are 1, 2, 3, 1
X[2][1], X[2][2], X[2][3] are |1-2| = 1, |2 - 3| = 1 and |3 - 1| = 2
X[3][1], X[3][2] are ? 1 − 1? = 0, ? 1 − 2? = 1.
X[4][1] = |0 - 1| = 1
So the answer is 1

To solve this, we will follow these steps −

Define a function calc(), this will take N, M,
cnt := 0
for initialize k := N, when k is non-zero, update k >>= 1, do:
   cnt := floor of (cnt + k)/2
for initialize k := M, when k is non-zero, update k >>= 1, do:
   cnt := floor of (cnt - k)/2
for initialize k := N - M, when k is non-zero, update k >>= 1, do:
   cnt := floor of (cnt - k)/2
return invert of cnt
From the main method, do the following
n := size of A
Define an array arr of size (n + 1)
for initialize i := 1, when i 
Example
Let us see the following implementation to get better understanding −
#include 
using namespace std;

int calc(int N, int M) {
   int cnt = 0;
   for (int k = N; k; k >>= 1)
      cnt += k >> 1;
   for (int k = M; k; k >>= 1)
      cnt -= k >> 1;
   for (int k = N - M; k; k >>= 1)
      cnt -= k >> 1;
   return !cnt;
}
string solve(vector A) {
   int n = A.size();
   vector arr(n + 1);
   for (int i = 1; i  A = { 1, 2, 3, 1 };
   cout 

Input
{ 1, 2, 3, 1 }
Output
1