Leave Me Alone by Stunna Cee Song Details

On July 26, 2018, the song "Leave Me Alone" was released by Stunna Cee. Leave Me Alone is considered explicit and may contain foul language. This song is about 3 minutes and 28 seconds, which is about the average length of a typical song. Because this song is the only song in Leave Me Alone and no other songs are present in the album, we classify this track as a single. Based on our statistics, Leave Me Alone's popularity is unknown right now. The overall tone is very danceable, especially with its high energy, which produces more of a euphoric, cheerful, or happy vibe.

BPM and Tempo

We consider the tempo marking of Leave Me Alone by Stunna Cee to be Allegro (fast, quick, and bright) because the track has a tempo of 150 BPM, a half-time of 75BPM, and a double-time of 300 BPM. Based on that, the speed of the song's tempo is fast. Activities such as, jogging or cycling, can go well with this song. The time signature for this track is 4/4.

Music Key

This song has a musical key of D Minor. Because this track belongs in the D Minor key, the camelot key is 7A. So, the perfect camelot match for 7A would be either 7A or 6B. While, a low energy boost can consist of either 7B or 8A. For moderate energy boost, you would use 4A and a high energy boost can either be 9A or 2A. However, if you are looking for a low energy drop, finding a song with a camelot key of 6A would be a great choice. Where 10A would give you a moderate drop, and 5A or 12A would be a high energy drop. Lastly, 10B allows you to change the mood.

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