Create EquiLeader.md

Author: yaseenshaik

EquiLeader.md

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+
+Task description
+A non-empty zero-indexed array A consisting of N integers is given.
+
+The leader of this array is the value that occurs in more than half of the elements of A.
+
+An equi leader is an index S such that 0 ≤ S < N − 1 and two sequences A[0], A[1], ..., A[S] and A[S + 1], A[S + 2], ..., A[N − 1] have leaders of the same value.
+
+For example, given array A such that:
+
+```
+    A[0] = 4
+    A[1] = 3
+    A[2] = 4
+    A[3] = 4
+    A[4] = 4
+    A[5] = 2
+```
+we can find two equi leaders:
+
+* 0, because sequences: (4) and (3, 4, 4, 4, 2) have the same leader, whose value is 4.
+* 2, because sequences: (4, 3, 4) and (4, 4, 2) have the same leader, whose value is 4.
+
+The goal is to count the number of equi leaders.
+
+Write a function:
+
+```
+function solution(A);
+```
+
+that, given a non-empty zero-indexed array A consisting of N integers, returns the number of equi leaders.
+
+For example, given:
+
+```
+    A[0] = 4
+    A[1] = 3
+    A[2] = 4
+    A[3] = 4
+    A[4] = 4
+    A[5] = 2
+```
+the function should return 2, as explained above.
+
+Assume that:
+
+* N is an integer within the range [1..100,000];
+* each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].
+
+Complexity:
+
+* expected worst-case time complexity is O(N);
+* expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).
+Elements of input arrays can be modified.
+
+```javascript
+function solution(A) {
+    var pos = 0;
+    var count = 0;
+
+    for (var i = 0; i < A.length; i++) {
+        if (A[pos] == A[i]) {
+            count++;
+        } else {
+            count--;
+            if (count == 0) {
+                pos = i;
+                count++;
+            }
+        }
+    }
+
+    var ret = 0;
+    var cand = A[pos];
+
+    var E = [];
+    var N = [];
+
+    var ec = 0;
+    var nc = 0;
+    for (var i = 0; i < A.length; i++) {
+        if (A[i] == cand) {
+            ec++;
+        } else {
+            nc++;
+        }
+        E[i] = ec;
+        N[i] = nc;
+    }
+
+    for (var i = 0; i < A.length; i++) {
+        if (E[i] > N[i] && ((nc - N[i]) < (ec - E[i]))) {
+            ret++;
+        }
+    }
+
+    return ret;
+}
+```