solution is that check if there is an entry in the array from i=2 to i=n-1; such that a[i]<a[i-1]&&a[i]<a[i+1]. If there is such a value ans is NO else answer is always "YES". This is an easy way to solve that problem.... OK................
hint to solve pillars problem educational codeforces round division 2.
solution is that check if there is an entry in the array from i=2 to i=n-1; such that a[i]<a[i-1]&&a[i]<a[i+1]. If there is such a value ans is NO else answer is always "YES". This is an easy way to solve that problem.... OK................
| Rev. | Lang. | By | When | Δ | Comment | |
|---|---|---|---|---|---|---|
| en2 |
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AM_I_Learning | 2019-07-23 13:50:22 | 23 | Tiny change: 'at problem' -> 'at problem.... OK................' (published) | |
| en1 |
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AM_I_Learning | 2019-07-23 13:49:36 | 291 | Initial revision (saved to drafts) |
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