https://github.com/avinandanbose/pop_push_in_stack_at_given_postion-through-array
Here is all about push and pop in STACK at given position.
https://github.com/avinandanbose/pop_push_in_stack_at_given_postion-through-array
Last synced: 6 days ago
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Here is all about push and pop in STACK at given position.
- Host: GitHub
- URL: https://github.com/avinandanbose/pop_push_in_stack_at_given_postion-through-array
- Owner: AvinandanBose
- License: mit
- Created: 2023-05-20T22:28:54.000Z (over 2 years ago)
- Default Branch: main
- Last Pushed: 2023-05-20T23:15:09.000Z (over 2 years ago)
- Last Synced: 2025-01-26T17:11:22.688Z (9 months ago)
- Size: 7.81 KB
- Stars: 1
- Watchers: 1
- Forks: 0
- Open Issues: 0
-
Metadata Files:
- Readme: README.md
- License: LICENSE
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README
𝑷𝑶𝑷 𝑷𝑼𝑺𝑯 𝑰𝑵 𝑺𝑻𝑨𝑪𝑲 𝑨𝑻 𝑮𝑰𝑽𝑬𝑵 𝑷𝑶𝑺𝑻𝑰𝑶𝑵(𝑰𝒎𝒑𝒍𝒆𝒎𝒆𝒏𝒕𝒆𝒅 𝑻𝒉𝒓𝒐𝒖𝒈𝒉 𝑨𝒓𝒓𝒂𝒚)
𝑷𝑼𝑺𝑯 𝑰𝑵 𝑺𝑻𝑨𝑪𝑲 𝑨𝑻 𝑮𝑰𝑽𝑬𝑵 𝑷𝑶𝑺𝑻𝑰𝑶𝑵(𝑰𝒎𝒑𝒍𝒆𝒎𝒆𝒏𝒕𝒆𝒅 𝑻𝒉𝒓𝒐𝒖𝒈𝒉 𝑨𝒓𝒓𝒂𝒚)
𝟏𝒔𝒕 𝒘𝒆 𝒉𝒂𝒗𝒆 𝒕𝒐 𝒄𝒉𝒆𝒄𝒌 𝒕𝒉𝒆 𝒂𝒓𝒓𝒂𝒚 𝒊𝒔 𝒇𝒖𝒍𝒍 𝒐𝒓 𝒏𝒐𝒕 𝒊𝒇 𝒕𝒉𝒆 𝒂𝒓𝒓𝒂𝒚 𝒊𝒔 𝒇𝒖𝒍𝒍,𝒊𝒕 𝒘𝒊𝒍𝒍 𝒓𝒆𝒕𝒖𝒓𝒏 𝑺𝒕𝒂𝒄𝒌 𝑶𝒗𝒆𝒓𝒇𝒍𝒐𝒘 𝒎𝒆𝒔𝒔𝒂𝒈𝒆.
𝑵𝒐𝒘 𝒔𝒖𝒑𝒑𝒐𝒔𝒆 𝒕𝒉𝒆 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏 𝒘𝒆 𝒉𝒂𝒗𝒆 𝒆𝒏𝒕𝒆𝒓𝒆𝒅 𝒊𝒔 𝒍𝒆𝒔𝒔 𝒕𝒉𝒂𝒏 𝟎 𝒐𝒓 𝒈𝒓𝒆𝒂𝒕𝒆𝒓 𝒕𝒉𝒂𝒏
𝒕𝒐𝒑 + 𝟏 𝒊. 𝒆. 𝒘𝒆 𝒉𝒂𝒗𝒆 𝒆𝒏𝒕𝒆𝒓𝒆𝒅 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏 = −𝟏 𝒐𝒓 𝒊𝒇 𝒕𝒉𝒆 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚 𝒐𝒇 𝒔𝒕𝒂𝒄𝒌 𝒊𝒔 𝟏𝟎
𝒂𝒏𝒅 𝒕𝒐𝒑 𝒊𝒔 𝒂𝒕 𝟔 ,𝒘𝒆 𝒄𝒂𝒏 𝒆𝒏𝒕𝒆𝒓 𝒐𝒓 𝒑𝒖𝒔𝒉 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝒂𝒕 𝟕 𝒃𝒖𝒕 𝒘𝒆 𝒄𝒂𝒏𝒏𝒐𝒕
𝒆𝒏𝒕𝒆𝒓 𝒐𝒓 𝒑𝒖𝒔𝒉 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝒂𝒕 𝟖 .
𝑯𝒆𝒏𝒄𝒆 𝒊𝒇 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏 > 𝒕𝒐𝒑 + 𝟏 ,𝒊𝒕 𝒘𝒊𝒍𝒍 𝒓𝒆𝒕𝒖𝒓𝒏 𝒊𝒏𝒗𝒂𝒍𝒊𝒅 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏.
𝑵𝒐𝒘, 𝒘𝒆 𝒘𝒊𝒍𝒍 𝒉𝒂𝒗𝒆 𝒔𝒂𝒎𝒆 𝒍𝒐𝒈𝒊𝒄 𝒘𝒉𝒆𝒏 𝒘𝒆 𝒂𝒓𝒆 𝒆𝒏𝒕𝒆𝒓𝒊𝒏𝒈 /𝒊𝒏𝒔𝒆𝒓𝒕𝒊𝒏𝒈
𝒆𝒍𝒆𝒎𝒆𝒏𝒕𝒔 𝒊𝒏 𝒂𝒓𝒓𝒂𝒚 𝒂𝒕 𝒂 𝒈𝒊𝒗𝒆𝒏 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏:
𝑺𝒂𝒚, 𝒔[𝟎] = 𝟏, 𝒔[𝟏] = 𝟐, 𝒔[𝟐] = 𝟑, 𝒔[𝟑] = 𝟒, 𝒔[𝟒] = 𝟓, 𝒔[𝟓] = 𝟔, 𝒔[𝟔] = 𝟕.
𝑨𝒏𝒅 𝒘𝒆 𝒉𝒂𝒗𝒆 𝒕𝒐 𝒆𝒏𝒕𝒆𝒓 𝟕 𝒂𝒕 𝒔[𝟑] .
𝑵𝒐𝒘, 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏 𝒊𝒔: 𝟑 𝒂𝒏𝒅 𝒕𝒐𝒑 𝒊𝒔 𝟕.
```Syntax
𝒔[𝟕] = 𝒔[𝟔] = 𝟕
𝒔[𝟔] = 𝒔[𝟓] = 𝟔
𝒔[𝟓] = 𝒔[𝟒] = 𝟓
𝒔[𝟒] = 𝒔[𝟑] = 𝟒
```
𝑵𝒐𝒘 𝒘𝒆 𝒉𝒂𝒗𝒆 𝒕𝒐 𝒆𝒏𝒕𝒆𝒓 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝒊𝒔 𝟕 𝒂𝒕 𝒔[𝟑].
𝒔[𝟑] = 𝟕.
𝒕𝒐𝒑 𝒊𝒔 𝟕 + 𝟏 = 𝟖 .
𝑵𝒐𝒘 𝒘𝒆 𝒘𝒊𝒍𝒍 𝒉𝒂𝒗𝒆 ∶ 𝒔[𝟎] = 𝟏, 𝒔[𝟏] = 𝟐, 𝒔[𝟐] = 𝟑, 𝒔[𝟑] = 𝟕,
𝒔[𝟒] = 𝟒, 𝒔[𝟓] = 𝟓, 𝒔[𝟔] = 𝟔, 𝒔[𝟕] = 𝟕.
𝑨𝒏𝒅 𝑻𝒐𝒑 𝒊𝒔 𝒂𝒕 𝒏𝒐𝒘: 𝟖.
𝑻𝒊𝒎𝒆 𝑪𝒐𝒎𝒑𝒍𝒆𝒙𝒊𝒕𝒚
𝑻𝒊𝒎𝒆 𝑪𝒐𝒎𝒑𝒍𝒆𝒙𝒊𝒕𝒚 ∶ 𝑶(𝟏) + 𝑶(𝟏) + 𝑶(𝒏) + 𝑶(𝟏) + 𝑶(𝟏)
= 𝑶(𝒏).
𝑺𝒑𝒂𝒄𝒆 𝑪𝒐𝒎𝒑𝒍𝒆𝒙𝒊𝒕𝒚
𝑺𝒑𝒂𝒄𝒆 𝒄𝒐𝒎𝒑𝒍𝒆𝒙𝒊𝒕𝒚 𝒐𝒇 𝒕𝒉𝒆 𝒂𝒃𝒐𝒗𝒆 𝒑𝒓𝒐𝒈𝒓𝒂𝒎 𝒊𝒔 𝑶(𝟏) 𝒂𝒔 𝒅𝒊𝒄𝒖𝒔𝒔𝒆𝒅
𝒆𝒂𝒓𝒍𝒊𝒆𝒓 .
𝑻𝒉𝒆 𝒔𝒑𝒂𝒄𝒆 𝒄𝒐𝒎𝒑𝒍𝒆𝒙𝒊𝒕𝒚 𝒐𝒇 𝒕𝒉𝒆 𝒑𝒖𝒔𝒉𝑨𝒕𝑷𝒐𝒔 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒊𝒔 𝑶(𝟏),
𝒘𝒉𝒊𝒄𝒉 𝒎𝒆𝒂𝒏𝒔 𝒊𝒕 𝒓𝒆𝒒𝒖𝒊𝒓𝒆𝒔 𝒂 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒂𝒎𝒐𝒖𝒏𝒕 𝒐𝒇 𝒆𝒙𝒕𝒓𝒂
𝒔𝒑𝒂𝒄𝒆 𝒓𝒆𝒈𝒂𝒓𝒅𝒍𝒆𝒔𝒔 𝒐𝒇 𝒕𝒉𝒆 𝒔𝒊𝒛𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒔𝒕𝒂𝒄𝒌.
𝑻𝒉𝒆 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒐𝒑𝒆𝒓𝒂𝒕𝒆𝒔 𝒅𝒊𝒓𝒆𝒄𝒕𝒍𝒚 𝒐𝒏 𝒕𝒉𝒆 𝒆𝒙𝒊𝒔𝒕𝒊𝒏𝒈 𝒔𝒕𝒂𝒄𝒌 𝒐𝒃𝒋𝒆𝒄𝒕 `𝒔𝒕`
𝒂𝒏𝒅 𝒅𝒐𝒆𝒔 𝒏𝒐𝒕 𝒖𝒔𝒆 𝒂𝒏𝒚 𝒂𝒅𝒅𝒊𝒕𝒊𝒐𝒏𝒂𝒍 𝒅𝒂𝒕𝒂 𝒔𝒕𝒓𝒖𝒄𝒕𝒖𝒓𝒆𝒔 𝒐𝒓
𝒅𝒚𝒏𝒂𝒎𝒊𝒄𝒂𝒍𝒍𝒚 𝒂𝒍𝒍𝒐𝒄𝒂𝒕𝒆 𝒎𝒆𝒎𝒐𝒓𝒚.
𝑰𝒕 𝒖𝒔𝒆𝒔 𝒂 𝒍𝒐𝒐𝒑 𝒕𝒐 𝒔𝒉𝒊𝒇𝒕 𝒆𝒍𝒆𝒎𝒆𝒏𝒕𝒔 𝒕𝒐 𝒕𝒉𝒆 𝒓𝒊𝒈𝒉𝒕 𝒇𝒓𝒐𝒎 𝒕𝒉𝒆
𝒈𝒊𝒗𝒆𝒏 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏, 𝒃𝒖𝒕 𝒕𝒉𝒆 𝒔𝒑𝒂𝒄𝒆 𝒖𝒔𝒆𝒅 𝒇𝒐𝒓 𝒕𝒉𝒆 𝒍𝒐𝒐𝒑
𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆 `𝒊` 𝒊𝒔 𝒏𝒆𝒈𝒍𝒊𝒈𝒊𝒃𝒍𝒆 𝒂𝒏𝒅 𝒅𝒐𝒆𝒔 𝒏𝒐𝒕 𝒅𝒆𝒑𝒆𝒏𝒅
𝒐𝒏 𝒕𝒉𝒆 𝒔𝒊𝒛𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒔𝒕𝒂𝒄𝒌.
𝑻𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆, 𝒕𝒉𝒆 𝒔𝒑𝒂𝒄𝒆 𝒄𝒐𝒎𝒑𝒍𝒆𝒙𝒊𝒕𝒚 𝒊𝒔 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕.
𝑷𝑶𝑷 𝑰𝑵 𝑺𝑻𝑨𝑪𝑲 𝑨𝑻 𝑮𝑰𝑽𝑬𝑵 𝑷𝑶𝑺𝑻𝑰𝑶𝑵(𝑰𝒎𝒑𝒍𝒆𝒎𝒆𝒏𝒕𝒆𝒅 𝑻𝒉𝒓𝒐𝒖𝒈𝒉 𝑨𝒓𝒓𝒂𝒚)
𝟏𝒔𝒕 𝒘𝒆 𝒉𝒂𝒗𝒆 𝒕𝒐 𝒄𝒉𝒆𝒄𝒌 𝒕𝒉𝒆 𝒂𝒓𝒓𝒂𝒚 𝒊𝒔 𝒆𝒎𝒑𝒕𝒚 𝒐𝒓 𝒏𝒐𝒕 𝒊𝒇 𝒕𝒉𝒆 𝒂𝒓𝒓𝒂𝒚 𝒊𝒔 𝒆𝒎𝒑𝒕𝒚,
𝒊𝒕 𝒘𝒊𝒍𝒍 𝒓𝒆𝒕𝒖𝒓𝒏 𝑺𝒕𝒂𝒄𝒌 𝑼𝒏𝒅𝒆𝒓𝒇𝒍𝒐𝒘 𝒎𝒆𝒔𝒔𝒂𝒈𝒆.
𝑵𝒐𝒘 𝒔𝒖𝒑𝒑𝒐𝒔𝒆 𝒕𝒉𝒆 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏 𝒘𝒆 𝒉𝒂𝒗𝒆 𝒆𝒏𝒕𝒆𝒓𝒆𝒅 𝒊𝒔 𝒍𝒆𝒔𝒔 𝒕𝒉𝒂𝒏 𝟎 𝒐𝒓 𝒈𝒓𝒆𝒂𝒕𝒆𝒓 𝒕𝒉𝒂𝒏
𝒕𝒐𝒑 𝒊. 𝒆.𝒘𝒆 𝒉𝒂𝒗𝒆 𝒆𝒏𝒕𝒆𝒓𝒆𝒅 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏 = −𝟏 𝒐𝒓 𝒊𝒇 𝒕𝒉𝒆 𝒄𝒂𝒑𝒂𝒄𝒊𝒕𝒚 𝒐𝒇 𝒔𝒕𝒂𝒄𝒌 𝒊𝒔 𝟏𝟎
𝒂𝒏𝒅 𝒕𝒐𝒑 𝒊𝒔 𝒂𝒕 𝟔 ,𝒏𝒐𝒘 𝒘𝒆 𝒄𝒂𝒏𝒏𝒐𝒕 𝒑𝒐𝒑 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝒂𝒕 𝟕 .
𝑯𝒆𝒏𝒄𝒆 𝒊𝒇 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏 > 𝒕𝒐𝒑 ,𝒊𝒕 𝒘𝒊𝒍𝒍 𝒓𝒆𝒕𝒖𝒓𝒏 𝒊𝒏𝒗𝒂𝒍𝒊𝒅 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏.
𝑵𝒐𝒘, 𝒘𝒆 𝒘𝒊𝒍𝒍 𝒉𝒂𝒗𝒆 𝒅𝒆𝒍𝒆𝒕𝒆 𝒂𝒏 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝒂𝒕 𝒂 𝒈𝒊𝒗𝒆𝒏 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏:
𝑺𝒂𝒚, 𝒔[𝟎] = 𝟏, 𝒔[𝟏] = 𝟐, 𝒔[𝟐] = 𝟑, 𝒔[𝟑] = 𝟒, 𝒔[𝟒] = 𝟓, 𝒔[𝟓] = 𝟔, 𝒔[𝟔] = 𝟕.
𝑨𝒏𝒅 𝒘𝒆 𝒉𝒂𝒗𝒆 𝒕𝒐 𝒅𝒆𝒍𝒆𝒕𝒆 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏 = 𝟑 .
𝒙 = 𝒔[𝟑] = 𝟒. 𝑨𝒏𝒅 𝒕𝒐𝒑 = 𝟕 𝒊. 𝒆. 𝒔[𝟔].
```Syntax
𝒔[𝟑] = 𝒔[𝟒] = 𝟓.
𝒔[𝟒] = 𝒔[𝟓] = 𝟔.
𝒔[𝟓] = 𝒔[𝟔] = 𝟕.
```
𝑻𝒐𝒑 = 𝑻𝒐𝒑 − 𝟏 = 𝟕 − 𝟏 = 𝟔.
𝑨𝒏𝒅 𝒑𝒐𝒑𝒑𝒆𝒅 𝒐𝒖𝒕 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 = 𝒔[𝟑] = 𝟒.
𝑵𝒐𝒘, 𝒘𝒆 𝒉𝒂𝒗𝒆 𝒔[𝟎] = 𝟏, 𝒔[𝟏] = 𝟐, 𝒔[𝟐] = 𝟑, 𝒔[𝟑] = 𝟓, 𝒔[𝟒] = 𝟔, 𝒔[𝟓] = 𝟕.
𝑻𝒊𝒎𝒆 𝑪𝒐𝒎𝒑𝒍𝒆𝒙𝒊𝒕𝒚
