CCC 2021 J4:Arrange books

https://cemc.uwaterloo.ca/contests/computing/2021/ccc/juniorEF.pdf

The problem is math problems.

Assume, L is the number of char L, M is the number of char M, S is the number of char S.

After arrangement,

the first block should be "L" from 0 to L-1

the second block should be "M" from L to L+M-1

the third block should be "S" from L+M to L+M+S -1.

Assume g1 has the L, M, S number in the first block.

Assume g2 has the L, M, S number in the first block.

Assume g3 has the L, M, S number in the first block.

Their two case:

1. L , M and S are not 0, then

Switch one step

switch two steps.

2. L, M and S has one 0, then:

result is the block one's another number.

Code:

SS= input()
L = 0
M = 0
S = 0
all = [[0,0,0]]
for x in SS:
    nL = all[-1][0]
    nM = all[-1][1]
    nS = all[-1][2]
    if x == "L":
        nL += 1
        L += 1
    elif x == "M":
        nM += 1
        M += 1
    else:
        nS += 1
        S += 1
    all.append([nL, nM, nS])
if L != 0 and M != 0 and S != 0:
    g1 = all[L]
    g2 = [all[L + M][0] - all[L][0], all[L + M][1] - all[L][1], all[L + M][2] - all[L][2]]
    g3 = [all[L + M + S][0] - all[L + M][0], all[L + M + S][1] - all[L + M][1], all[L + M + S][2] - all[L + M][2]]
    step = 0
    step += min(g1[1], g2[0]) + min(g1[2], g3[0])
    g1[1] -= min(g1[1], g2[0])
    g2[0] -= min(g1[1], g2[0])
    g1[2] -= min(g1[2], g3[0])
    g3[0] -= min(g1[2], g3[0])
    step += min(g2[2], g3[1])
    g2[2] -= min(g2[2], g3[1])
    g3[1] -= min(g2[2], g3[1])
    step += g1[1]*2 + g1[2]*2
    print(step)

else:
    if L == 0:
        if M != 0 and S != 0:
            print(all[M][2])
        else:
            print(0)
    elif M == 0:
        if L != 0 and S != 0:
            print(all[L][2])
        else:
            print(0)
    else:
        if L != 0 and M != 0:
            print(all[M][1])
        else:
            print(0)