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................
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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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