Approach:
- First got the max num from array.
- Using the above max num, got the second max num
- If the 2nd Largest num is present, I have returned it.
- If such 2nd largest num is absent, I have returned -1.
int Solution::solve(vector<int> &A) { int max_num = INT_MIN; int max_sec = INT_MIN; int n = A.size(); for(int i = 0; i < n; i++) { max_num = max(max_num, A[i]); } for(int i = 0; i < n; i++) { if(A[i] != max_num) { max_sec = max(max_sec, A[i]); } } if(max_sec != INT_MIN) { return max_sec; } else{ return -1; } } 
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