# 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[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 to X are 1, 2, 3, 1
X, X, X are |1-2| = 1, |2 - 3| = 1 and |3 - 1| = 2
X, X are ∣ 1 − 1∣ = 0, ∣ 1 − 2∣ = 1.
X = |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 < n, update (increase i by 1), do:
arr[i - 1] = |A[i] - A[i - 1]|
(decrease n by 1)
hh := 1, pd := 0, ck := 0
for initialize i := 0, when i < n, update (increase i by 1), do:
if arr[i] is non-zero, then:
if arr[i] is same as 1, then:
hh := 0
pd := pd XOR calc(n - 1, i)
Otherwise
ck := ck XOR calc(n - 1, i)
ck := ck AND hh
if pd XOR ck is non-zero, then:
return "1" + hh
return "0"

## Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
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<int> A) {
int n = A.size();
vector<int> arr(n + 1);
for (int i = 1; i < n; i++) {
arr[i - 1] = abs(A[i] - A[i - 1]);
}
--n;
bool hh = 1, pd = 0, ck = 0;
for (int i = 0; i < n; i++)
if (arr[i]) {
if (arr[i] == 1)
hh = 0, pd ^= calc(n - 1, i);
else
ck ^= calc(n - 1, i);
}
ck &= hh;
if (pd ^ ck)
return "1" + hh;
return "0";
}
int main(){
vector<int> A = { 1, 2, 3, 1 };
cout << solve(A) << endl;
}

## Input

{ 1, 2, 3, 1 }

## Output

1