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Joined 2026-09-21 Message
Incoming links: Two-way array partition
The original page permits , but also indexes . The meaningful pivot-position contract is the stronger , with and . Implement this two-way array partition by reserving the last entry as pivot:The loop invariant isThe segment's multiset is preserved and entries outside are unchanged. The unclassified part is . A smaller entry is swapped into the left group; if , the displaced entry belongs to the greater-or-equal group. Otherwise the entry simply joins that group. The loop variant is .
PROCEDURE Partition(l, r: INTEGER): INTEGER;
VAR pivot, i, j, tmp: INTEGER;
BEGIN
ASSERT(l < r);
pivot := a[r-1];
i := l; j := l;
WHILE j < r-1 DO
IF a[j] < pivot THEN
tmp := a[i]; a[i] := a[j]; a[j] := tmp;
i := i+1
END;
j := j+1
END;
tmp := a[i]; a[i] := a[r-1]; a[r-1] := tmp;
RETURN i
END Partition;Store the current multiset in an unsorted prefix The loop invariant is , the complete array has its original multiset, every discarded entry on the left is at most every active entry, and every active entry is at most every discarded entry on the right. Thus the desired global order statistic is the local one of rank . The two-way array partition result either identifies it or excludes the pivot and the irrelevant side. The positive loop variant strictly decreases unless the procedure returns.
a[0..len); Add appends exactly as in part (a). The coupling invariant is equality of length and multiset with abstract xs, rather than equality of insertion order. Get applies quickselect:PROCEDURE Get(k: INTEGER): INTEGER;
VAR l, r, m: INTEGER;
BEGIN
ASSERT((0 <= k) & (k < len));
l := 0; r := len;
LOOP
m := Partition(l, r);
IF k = m THEN RETURN a[m]
ELSIF k < m THEN r := m
ELSE l := m+1
END
END
END Get;Balanced pivots give , whose geometric series is . More generally, choosing a uniformly random pivot before each call, by swapping a random active entry into
a[r-1], gives expected running time on distinct keys: with constant probability, the pivot is in the middle half and the retained side has size at most . Expected work over these shrinking stages is a convergent geometric series.The particular last-entry pivot in the displayed code has no unconditional expected guarantee without an input distribution. Its worst case repeatedly discards one entry. Duplicate keys require further care: on an all-equal segment, the strict partition always returns , so selecting the last rank costs even with randomized pivot positions. A three-way array partitioning that groups all keys equal to the pivot restores the usual randomized linear expectation with duplicates; alternatively the stated expectation assumes distinct keys. This is a qualification to the bound, not to the algorithm correctness of the displayed code.Let The loop invariant is , every entry of is less than , every entry of is at least , and the segment's multiset is unchanged. Entries outside are unchanged. When
a be the given enclosing array. A two-way array partition needs only one scan, and does not require the pivot to occur in the array:def partition(l: Int, r: Int, p: Int): Int = {
require(0 <= l && l <= r && r <= a.length)
var k = l
var j = l
while (j < r) {
if (a(j) < p) {
val t = a(k)
a(k) = a(j)
a(j) = t
k += 1
}
j += 1
}
k
}a(j)<p, swapping it into position k extends the low part; if k<j, the displaced entry was in the high part and remains there after j advances. If k==j, the self-swap is harmless. The other branch simply extends the high part. The loop variant is . At exit , so the returned has the required separation property, including empty segments and arbitrarily many duplicate entries. The time complexity is and auxiliary space complexity is .Keep a deficient interval of possible values and only the array segment containing occurrences from that interval. This is missing-value search by deficient value intervals:The precondition is that every original entry lies in . The loop invariant istogether with: every active entry lies in , and every original occurrence with a value in that interval is in the active segment. The whole array remains a permutation of its initial contents. Initially these assertions follow from the precondition and .
def missing(): Int = {
val n = a.length
var lo = 0L
var hi = n.toLong + 1L
var l = 0
var r = n
while (hi - lo > 1L) {
val p = lo + (hi - lo) / 2L
val k = partition(l, r, p.toInt)
if ((k - l).toLong < p - lo) {
hi = p
r = k
} else {
lo = p
l = k
}
}
lo.toInt
}After array partitioning, the low subinterval has occurrences and possible values. If it is deficient, retaining it preserves the loop invariant. Otherwise , while the original active count is strictly less than ; subtraction givesThus the high subinterval is deficient and retaining it also preserves the loop invariant. Discarded occurrences lie outside the new candidate interval. The loop variant is the positive integer , which strictly decreases because the midpoint is strictly internal. At exit the interval has one possible value and fewer than one occurrence, so
lo is absent from the entire original array. For the loop is skipped and zero is returned. The Long value bounds avoid overflowing ; every internal pivot is at most , so its checked mathematical range fits Int.For the time complexity, let be the interval width and its active occurrence count before iteration . The loop invariant gives , and binary search gives . Each two-way array partition costs . Until the width reaches one, the widths are bounded by , so the total scan work is bounded by a geometric series plus loop overhead: overall, hence for positive . Partitioning the full array again at every iteration would instead lose this bound. Auxiliary space complexity is .