# Find longest sequence of zeros in list

How it is possible to find the length of the longest sequence of zeros in list

{1,0,0,1,1,0,0,0,0,1,0,0,0,1,1,0,0,1}


which is equal to 4 in the sample list above

• Another way: SequenceCases[seq, {p : Repeated} :> Length[{p}]] // Max (too similar to others, and to docs, to post as answer). – user1066 Jul 30 '18 at 9:29

Assuming it is a binary sequence, one way is to look at all the differences in the positions of the 1s and take the largest:

seq = {1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1};
Max[Differences[Position[seq, 1]]] - 1
4


As pointed out in the comments by jkuczm and Hector, this will fail if the longest sequence of zeros occurs at either the start or the end. This can be fixed by surrounding the sequence with 1s:

Max[Differences[Position[Flatten@{1, seq, 1}, 1]]] - 1

• I must say this is an awful lot faster than using patterns! 0.000115 seconds for this vs 0.221 for my implementation of patterns in a list of 500 integers. – Carl Lange Jul 29 '18 at 23:20
• There are more assumptions for this to work. You need at least two 1s and longest sequence of zeros can't start at first element. So it'll fail e.g. for {1, 0, 0, 0} or {0, 0, 0, 1, 0, 1}. – jkuczm Jul 30 '18 at 9:40
• The assumptions are easy to satisfy: Max[Differences[Position[Join[{1,1},seq], 1]]] - 1 – Hector Jul 30 '18 at 10:50
• @Hector it still fails for {1, 0, 0, 0}. I guess that correct assumption is not that "longest sequence of zeros can't start at first element" as I've written before, but that sequence of zeros, to be counted, should be surrounded by 1s. So 1 should be added at beginning and at end as in current version of answer by kglr: Max[Differences[Position[Join[{1}, seq, {1}], 1]]] - 1. – jkuczm Jul 30 '18 at 14:38
Max@FoldList[If[#2 == 0, #1 + 1, 0] &, 0, seq]


If you need speed and are only using packable arrays you could use a compiled function. Function generator returning compiled functions, performing desired operation can look like this:

compileLongestSeqLen // ClearAll
compileLongestSeqLen[
{type_, rank_Integer /; rank >= 1}, inSeqQ_, opts : OptionsPattern@Compile
] := Compile[{{list, type, rank}},
Module[{len = 0, longest = 0},
Do[
If[inSeqQ@x,
++len;
(* else *),
If[len > longest, longest = len];
len = 0;
]
,
{x, list}
];
If[len > longest, longest = len];
longest
],
opts
]


Let's compile function finding length of longest sequence of zeros in list of integers:

jkuczm = compileLongestSeqLen[{_Integer, 1}, # === 0 &,
RuntimeOptions -> "Speed", CompilationTarget -> "C",
RuntimeAttributes -> {Listable}, Parallelization -> True
]


To compare efficiency let's define functions from other answers:

alan = Max@FoldList[If[#2 == 0, #1 + 1, 0] &, 0, #] &;
billSHect1 = Max[Differences[Position[Flatten@{1, #, 1}, 1]]] - 1 &;
billSHect2 = Max[Differences[Position[Join[{1}, #, {1}], 1]]] - 1 &;
carlL = # /. {___, a : Longest[Repeated], ___} :> Length[{a}] &;
chyanog1 = GroupBy[Split[#], First -> Length, Max] &;
chyanog2 = (# // Split // Select[MemberQ] // Map[Length] // Max) &;
joe = (SortBy[Split[#], Length] // Last // Length) &;
kglr1 = Max[Differences@PositionIndex[Join[{1}, #, {1}]]] - 1 &;
kglr2 = Max[Length /@ Split@Accumulate[Join[{1}, #, {1}]]] - 1 &;
maxZeros[a_List] := Max[Append[# - 1, Length@a] - Prepend[#, 0]] &@SparseArray[a]["AdjacencyLists"]
okkes = (Length /@ Split[#] // Max) &;


Timings for single large list:

SeedRandom@0
test = RandomInteger[{0, 1}, 10^6];

(*carlLRes = carlL@test; // MaxMemoryUsed // RepeatedTiming*)   (*too slow*)
chyanog2Res = chyanog2@test; // MaxMemoryUsed // RepeatedTiming (*{0.62,  44510360}*)
(*joeRes = joe@test; // MaxMemoryUsed // RepeatedTiming*)       (*incorrect results*)
chyanog1Res = chyanog1@test; // MaxMemoryUsed // RepeatedTiming (*{0.36,  64019000}*)
billSRes1 = billSHect1@test; // MaxMemoryUsed // RepeatedTiming (*{0.33, 119935248}*)
billSRes2 = billSHect2@test; // MaxMemoryUsed // RepeatedTiming (*{0.31, 119934952}*)
kglr2Res = kglr2@test; // MaxMemoryUsed // RepeatedTiming       (*{0.26,  75941464}*)
(*okkesRes = okkes@test; // MaxMemoryUsed // RepeatedTiming*)   (*incorrect results*)
kglr1Res = kglr1@test; // MaxMemoryUsed // RepeatedTiming       (*{0.11,  44568024}*)
alanRes = alan@test; // MaxMemoryUsed // RepeatedTiming         (*{0.063, 40008744}*)
mrWizRes = maxZeros@test; // MaxMemoryUsed // RepeatedTiming    (*{0.0273 17126232}*)
jkuczmRes = jkuczm@test; // MaxMemoryUsed // RepeatedTiming     (*{0.0047,      56}*)

(*carlLRes ===*) chyanog2Res === (*joeRes ===*) chyanog1Res === billSRes1 === billSRes2 === kglr2Res === (*okkesRes ===*) kglr1Res === alanRes === mrWizRes === jkuczmRes
(* True *)


Timings for multiple lists:

SeedRandom@0
test = RandomInteger[{0, 1}, {100, 10^5}];

(*carlLRes=carlL/@test;//MaxMemoryUsed//RepeatedTiming*)           (*too slow*)
chyanog2Res = chyanog2 /@ test; // MaxMemoryUsed // RepeatedTiming (*{6.3,  84491416}*)
(*joeRes = joe /@ test; // MaxMemoryUsed // RepeatedTiming*)       (*incorrect results*)
chyanog1Res = chyanog1 /@ test; // MaxMemoryUsed // RepeatedTiming (*{3.46, 86460488}*)
billSRes1 = billSHect1 /@ test; // MaxMemoryUsed // RepeatedTiming (*{3.29, 92106568}*)
kglr2Res = kglr2 /@ test; // MaxMemoryUsed // RepeatedTiming       (*{2.68, 87630888*)
(*okkesRes = okkes /@ test; // MaxMemoryUsed // RepeatedTiming*)   (*incorrect results*)
kglr1Res = kglr1 /@ test; // MaxMemoryUsed // RepeatedTiming       (*{1.12, 84780440}*)
alanRes = alan /@ test; // MaxMemoryUsed // RepeatedTiming         (*{0.379, 2409680}*)
billSRes2 = billSHect2 /@ test; // MaxMemoryUsed // RepeatedTiming (*{0.24,  2400320}*)
mrWizRes = maxZeros /@ test; // MaxMemoryUsed // RepeatedTiming    (*{0.22, 81755512}*)
jkuczmRes = jkuczm@test; // MaxMemoryUsed // RepeatedTiming        (*{0.030, 1602456}*)

(*carlLRes ===*) chyanog2Res === (*joeRes ===*) chyanog1Res === billSRes1 === kglr2Res === (*okkesRes ===*) kglr1Res === alanRes === billSRes2 === mrWizRes === jkuczmRes
(* True *)


This is relatively straightforward using patterns.

l /. {___, a : Longest[Repeated], ___} :> Length[{a}]


That is, name list a the Longest list of Repeated zeroes bounded on either side by anything not a zero (the ___ matches none or more of anything), and then get the Length of that list a.

GroupBy[Split[seq], First -> Length, Max]


or

seq // Split // Select[MemberQ] // Map[Length] // Max

f1 = Max[Differences @ PositionIndex[Join[{1}, #, {1}]]] - 1&;
f2 = Max[Length /@ Split @ Accumulate[Join[{1}, #, {1}]]] - 1 &;

list1 = {1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1};
list2 = {0, 0, 0, 0,  0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1};

f1 @ list1


4

f1 @ list2


6

f2 @ list1


4

f2 @ list2


6

• First implementation, as in bill s answer, will fail if there are less than two 1s. Both will fail if longest sequence of zeros starts with first element. – jkuczm Jul 30 '18 at 9:47
• Thank you again @jkuczm. Was able to salvage both with a minor tweak. – kglr Jul 30 '18 at 10:54
• I think that f1 is now correct. f2 fails if there's 0 as first element, but it's not part of longest sequence of 0s, e.g. f2@{0, 1, 0, 0} (* 3 *). – jkuczm Jul 30 '18 at 14:47

I'm not a Mathematica user, but in several other languages I generalized this to (run-length encoding)

rle(x==0)


which eliminates the requirement that x be only 1 or zero.

(EDIT by corey979)

As per Karsten's comment, the rle can be implemented in MMA as

RunLengthEncode[x_List] := {First[#], Length[#]}& /@ Split[x]


At least in Mathematica 10.1 we can improve on bill s's solution by an order of magnitude using SparseArray Properties as I did for Find subsequences of consecutive integers inside a list.

My proposal:

maxZeros[a_List] :=
Max[Append[# - 1, Length@a] - Prepend[#, 0]] & @ SparseArray[a]["AdjacencyLists"]


Timings:

Needs["GeneralUtilities"]

f1[a_] := Max[Differences[Position[Flatten@{1, a, 1}, 1]]] - 1

BenchmarkPlot[{f1, maxZeros}, RandomInteger[1, #] &]
` 