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Commit cad0ced

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Let us take the minimum and maximum elements of this array
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A[i] - A[1] + A[n] - A[i] = A[n] - A[i]
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-----
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void solve()
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{
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int no_of_elements;
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cin >> no_of_elements;
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vector <int> A(no_of_elements + 1);
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for(int i = 1; i <= no_of_elements; i++)
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{
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cin >> A[i];
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}
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int minimum = 1, maximum = 1;
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for(int i = 1; i <= no_of_elements; i++)
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{
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if(A[i] >= A[maximum])
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{
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maximum = i;
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}
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else if(A[i] < A[minimum])
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{
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minimum = i;
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}
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}
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cout << minimum << " " << maximum << "\n";
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}
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Notice that we can never change a 0 or 1
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If there is a 0 or 1, it will remain 0 or 1 throughout all the operations that we do.
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The next observation to make is that we can set every element = 0
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We can start with the largest element and make all elements = 0 by choosing x = max of the current array
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This works, except when there is a 1 in the array
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-----
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When there is a 1 in the array, we can follow a similar algorithm and set all elements = 1,
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by choosing X = Array Max - 1 at each step
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The only exception is when we have 2 consecutive elements.
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When we have 2 consecutive elements, we can never make both elements = 1
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We will finish at a situation where m = 1 and m + 1 = 2 or m = 0 and m + 1 = 2
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In both cases, we can not make 0 or 2 = 1 by any operation
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Now, it is impossible.
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------
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It is impossible when there is a 1 in the array and there are consecutive elements in the array
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Otherwise, we can make all elements = 0 or 1 depending on whether the array has a 1 or not.
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-----
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void solve()
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{
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int no_of_elements;
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cin >> no_of_elements;
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vector <long long> A(no_of_elements + 1);
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for(int i = 1; i <= no_of_elements; i++)
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{
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cin >> A[i];
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}
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vector <int> present(3, false);
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for(int i = 1; i <= no_of_elements; i++)
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{
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if(A[i] <= 2)
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{
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present[A[i]] = true;
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}
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}
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int possible = true;
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if(present[1])
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{
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sort(A.begin(), A.end());
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for(int i = 1; i + 1 <= no_of_elements; i++)
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{
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if(A[i] + 1 == A[i + 1])
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{
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possible = false;
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break;
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}
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}
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}
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cout << (possible ? "Yes" : "No") << "\n";
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}

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